我的hihocoder存代码
2024-04-10 07:39:33
hihocoder的教程挺好理解，这里存代码。
1014 Trie树
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define eps 10^(-6)
#define Q_CIN ios::sync_with_stdio(false);
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);
#define MOD 10009
#define debug(x) cout<<#x<<":"<<(x)<<endl;
#define lson i<<1,l,m
#define rson i<<1|1,m+1,rstruct tree
{tree *next[27];int cnt;tree(){cnt=0;CLR(next,NULL);}
};
int n,m;
void add(tree *p,char *s1)
{int i=0;while(s1[i]){int k=s1[i]-'a';if(!p->next[k])p->next[k]=new tree();p=p->next[k];p->cnt++;i++;}
}
int query(tree *p,char *s1)
{int i=0;while(s1[i]){int k=s1[i]-'a';if(!p->next[k]) return 0;p=p->next[k];i++;}return p->cnt;
}int main()
{
// freopen("1.in","r",stdin);int n,m;char s1[30];while(~scanf("%d%*c",&n)){tree *p=new tree();for(int i=1;i<=n;i++){scanf("%s",s1);add(p,s1);}scanf("%d%*c",&m);for(int i=1;i<=m;i++){scanf("%s",s1);printf("%d\n",query(p,s1));}}return 0;
}
1015 KMP算法
#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);int Next[10005];
string s1,s2;
void getNext(string s1)
{int i=-1,j=0;int len=s1.length();Next[0]=-1;while(j<len){if(i==-1||s1[i]==s1[j])Next[++j]=++i;elsei=Next[i];}}int kmp(string s1,string s2) //s2里找s2,可重叠
{int i=0,j=0;int len1=s1.length();int len2=s2.length();int cnt=0;for(int i=0;i<len1;i++){while(j!=-1&&s1[i]!=s2[j])j=Next[j];if(j==-1||s1[i]==s2[j])j++;if(j==len2)cnt++;}return cnt;
}
int main()
{REint n;cin>>n;for(int i=0;i<n;i++){cin>>s1>>s2;getNext(s1);cout<<kmp(s2,s1)<<endl;}
}1032 最长回文子串
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define eps 10^(-6)
#define Q_CIN ios::sync_with_stdio(false);
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);
#define MOD 10009
#define bug(x) cout<<#x<<":"<<(x)<<endl;char s1[1000020],s2[2000100];
int len[1000000*2+8];int init()
{int len=strlen(s1),cnt=0;REP(i,len){s2[cnt++] = '#';s2[cnt++] = s1[i];}s2[cnt++]='#';s2[cnt]='\0';return strlen(s2);
}int Manacher()
{int n=init();int pos=0,mx=0,ans=1;FOR(i,0,n-1){if(mx>i){len[i]=min(len[pos*2-i],mx-i);}else{len[i]=1;}while(i-len[i]>=0&&s2[i+len[i]]==s2[i-len[i]]) {len[i]++;}if(len[i]+i>mx){mx=len[i]+i;pos=i;}ans=max(ans,len[i]);}return ans-1;
}int main()
{
// REint t;scanf("%d%*c",&t);while(t--){scanf("%s",s1);printf("%d\n",Manacher());}return 0;
}
1050 树中的最长路
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) );
#define MOD 10009
#define bug(x) cout<<#x<<":"<<(x)<<endl;vector<vector<int> > v;
int src,len;
// 任意一个点dfs求距离它最远的点，再从该点dfs求距离该点最远的点，则这次距离就是了
void dfs(int u,int depth,int father){ //father:父结点if( depth > len){len = depth;src = u;}for (int i = 0; i < v[u].size(); ++i){if(father != v[u][i]) //不能走回头路dfs(v[u][i],depth+1,u);}
}int main(){int n,a,b;cin>>n;v.resize(n+1); //!!!!!!!!!!!!!for (int i = 0; i < n-1; ++i){cin>>a>>b;v[a].push_back(b);v[b].push_back(a);}len = 0;dfs(1,0,0);len = 0;dfs(src,0,0);cout<<len<<endl;return 0;
}1093 最短路径·三：SPFA算法
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define eps 10^(-6)
#define Q_CIN ios::sync_with_stdio(false);
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);
#define MOD 10009
#define debug(x) cout<<#x<<":"<<(x)<<endl;
const int maxn=1e5+5;
const int maxm=1e6+5;
const int inf=1e3*maxm;
int inq[maxn],head[maxn],dis[maxn];
struct Edge
{int v,w,next;
}edge[maxm*2];
int cnt;
void add_edge(int u,int v,int w)
{edge[cnt].v=v;edge[cnt].w=w;edge[cnt].next=head[u];head[u]=cnt++;
}
void init(int n)
{cnt=0;CLR(head,-1);CLR(inq,0);FOR(i,1,n)dis[i]=inf;
}
int spfa(int s,int t)
{queue<int>q;dis[s]=0;inq[s]=1;q.push(s);while(!q.empty()){int u=q.front();q.pop();inq[u]=0;for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].v;int w=edge[i].w;if(dis[v]>dis[u]+w){dis[v]=dis[u]+w;if(!inq[v]){inq[v]=1;q.push(v);}}}}return dis[t];
}
int main()
{
// REint n,m,s,t,a,b,c;while(cin>>n>>m>>s>>t){init(n);REP(i,m){cin>>a>>b>>c;add_edge(a,b,c);add_edge(b,a,c);}cout<<spfa(s,t)<<endl;}return 0;
}
1121 二分图一•二分图判定
给定的图可能不联通，所以一次搜索不行，如例子：
4 2
1 2
3 4
应为Correct
BFS
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define INF 0x7FFFFFFF
#define INT_MIN -(1<<31)
#define eps 10^(-6)
#define Q_CIN ios::sync_with_stdio(false);
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);
#define MOD 10009
#define debug(x) cout<<#x<<":"<<(x)<<endl;
const int maxn=10050;
vector<int>v[maxn];
int vis[maxn];
int n;
int match(int s)
{queue<int>q;q.push(s);vis[s]=1;int flag=0;while(!q.empty()){int pre=q.front();q.pop();for(int i=0;i<v[pre].size();i++){int x=v[pre][i];if(vis[x]==-1){vis[x]=!vis[pre];q.push(x);}else if(vis[x]==vis[pre])return 0;}}return 1;
}
int main()
{
// REint t,m,a,b;cin>>t;while(t--){cin>>n>>m;FOR(i,1,n)v[i].clear();CLR(vis,-1);REP(i,m){cin>>a>>b;v[a].push_back(b);v[b].push_back(a);}int ok=1;FOR(i,1,n){if(vis[i]==-1&&!match(i)){ok=0;cout<<"Wrong"<<endl;break;}}if(ok)cout<<"Correct"<<endl;}return 0;
}
DFS
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define INF 0x7FFFFFFF
#define INT_MIN -(1<<31)
#define eps 10^(-6)
#define Q_CIN ios::sync_with_stdio(false);
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);
#define MOD 10009
#define debug(x) cout<<#x<<":"<<(x)<<endl;
const int maxn=10050;
vector<int>v[maxn];
int vis[maxn];
int n;
int dfs(int s)
{for(int i=0;i<v[s].size();i++){int x=v[s][i];if(vis[x]==-1){vis[x]=!vis[s];if(!dfs(x)) return 0;}else if(vis[x]==vis[s])return 0;}return 1;
}
int main()
{REint t,m,a,b;cin>>t;while(t--){cin>>n>>m;FOR(i,1,n)v[i].clear();CLR(vis,-1);REP(i,m){cin>>a>>b;v[a].push_back(b);v[b].push_back(a);}int ok=1;FOR(i,1,n){if(vis[i]==-1){vis[i]=1; //不能放进dfs里if(!dfs(i)) {ok=0;cout<<"Wrong"<<endl;break;}}}if(ok)cout<<"Correct"<<endl;}return 0;
}
1122 二分图二 分图最大匹配之匈牙利算法
注意可能有多个联通块
#include <cstdio>
#include <cstring>
#include <queue>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);
const int maxn=1005;
const int maxm=5005;int mat[maxn][maxn];
int arr[maxn][maxn];
int vis[maxn];
int match[maxn];
int team[maxn];int n,m;
int cnt;int can(int u)
{for(int v=1;v<=n;v++){if(mat[u][v]&&!vis[v]){vis[v]=1;if(match[v]==-1||can(match[v])){match[v]=u;match[v]=u;return 1;}}}return 0;
}
void bfs(int u)
{queue<int>q;q.push(u);team[u]=0;while(!q.empty()){int x=q.front();q.pop();for(int i=1;i<=n;i++){if(arr[x][i]&&team[i]==-1){team[i]=!team[x];q.push(i);}}}
}
int main()
{
// REint a,b;while(cin>>n>>m){cnt=0;CLR(mat,0);CLR(arr,0);CLR(team,-1);REP(i,m){cin>>a>>b;if(a!=b)arr[a][b]=arr[b][a]=mat[a][b]=mat[b][a]=1;}for(int i=1;i<=n;i++){if(team[i]==-1){bfs(i);}}CLR(match,-1);FOR(i,1,n){CLR(vis,0);if(!team[i]&&can(i))cnt++;}cout<<cnt<<endl;}return 0;
}二分图三 二分图最小点覆盖和最大独立集
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define eps 10^(-6)
#define Q_CIN ios::sync_with_stdio(false);
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);
#define MOD 10009
#define debug(x) cout<<#x<<":"<<(x)<<endl;
#define lson i<<1,l,m
#define rson i<<1|1,m+1,rint n,m;
int link[1005],vis[1005];
int mat[1005][1005];
int team[1005];
int can(int u)
{for(int v=1;v<=n;v++){if(mat[u][v]&&!vis[v]){vis[v]=1;if(link[v]==-1||can(link[v])){link[v]=u;return 1;}}}return 0;
}
void bfs(int x)
{queue<int>q;q.push(x);team[x]=0;while(!q.empty()){int t=q.front();q.pop();for(int v=1;v<=n;v++)if(mat[t][v]&&team[v]==-1){team[v]=!team[t];q.push(v);}}
}
int main()
{
// freopen("1.in","r",stdin);int a,b;while(~scanf("%d%d",&n,&m)){memset(mat,0,sizeof(mat));memset(link,-1,sizeof(link));memset(team,-1,sizeof(team));int cnt=0;for(int i=1;i<=m;i++){scanf("%d%d",&a,&b);mat[a][b]=mat[b][a]=1;}for(int i=1;i<=n;i++)if(team[i]==-1)bfs(i);for(int i=1;i<=n;i++){memset(vis,0,sizeof(vis));if(!team[i]&&can(i)) cnt++;}printf("%d\n%d\n",cnt,n-cnt);}return 0;
}
RMQ-ST算法
cin超时
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define eps 10^(-6)
#define Q_CIN ios::sync_with_stdio(false);
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);
#define MOD 10009
#define debug(x) cout<<#x<<":"<<(x)<<endl;
#define lson i<<1,l,m
#define rson i<<1|1,m+1,r
int arr[1000005],dp[1000005][20];
int n,m;
void init()
{for(int i=1;i<=n;i++)dp[i][0]=arr[i];for(int j=1;(1<<j)<=n;j++){for(int i=1;i+(1<<j)-1<=n;i++){dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}}
}
int query(int l,int r)
{int k=(int)log2(r-l+1.0);return min(dp[l][k],dp[r-(1<<k)+1][k]);
}
int main()
{
// freopen("1.in","r",stdin);int l,r;scanf("%d",&n);{for(int i=1;i<=n;i++)scanf("%d",&arr[i]);scanf("%d",&m);init();for(int i=1;i<=m;i++){scanf("%d%d",&l,&r);printf("%d\n",query(l,r));}}return 0;
}
1077 RMQ问题再临-线段树
单点更新，区间查询
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <map>
#include <queue>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define clr( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);
#define MOD 10009
#define bug(x) cout<<#x<<":"<<(x)<<endl;
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
const int maxn = 1e6 + 5;
const int maxm = 1e6 + 5;
const int inf = 0x3f3f3f3f;int val[maxn << 2];
void build(int rt, int l, int r) {if (l == r) {scanf("%d",&val[rt]);} else {int m = (l + r) >> 1;build(lson);build(rson);val[rt] = min(val[rt << 1], val[rt << 1 | 1]);}
}
void update(int rt, int l, int r, int pos, int num) {if (l == r && r == pos) {val[rt] = num;} else {int m = (l + r) >> 1;if (pos <= m)update(lson, pos, num);elseupdate(rson, pos, num);val[rt] = min(val[rt << 1], val[rt << 1 | 1]);}
}int query(int rt, int l, int r, int q1, int q2) {if (q1 <= l && r <= q2) {return val[rt];}int m = (l + r) >> 1, t1 = inf, t2 = inf;if (q1 <= m)t1 = query(lson, q1, q2);if (q2 > m)t2 = query(rson, q1, q2);return min(t1, t2);
}int main() {int a, b, c, m, n;scanf("%d",&n);build(1, 1, n);scanf("%d",&m);while (m--) {scanf("%d%d%d",&a,&b,&c);if (a) {update(1, 1, n, b, c);} else {printf("%d\n", query(1, 1, n, b, c));}}return 0;
}
1078 线段树的区间修改线段树
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define eps 10^(-6)
#define Q_CIN ios::sync_with_stdio(false);
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);
#define MOD 10009
#define debug(x) cout<<#x<<":"<<(x)<<endl;
#define lson i<<1,l,m
#define rson i<<1|1,m+1,r
int sum[1000005<<2],add[1000005<<2];
int n,m;
void build(int i,int l,int r)
{add[i]=0;if(l==r){scanf("%d",&sum[i]);return;}int m=(l+r)/2;build(lson);build(rson);sum[i]=sum[i<<1]+sum[i<<1|1];
}void pushdown(int i,int m)
{add[i<<1]=add[i<<1|1]=add[i];sum[i<<1]=add[i]*(m-m/2);sum[i<<1|1]=add[i]*(m/2);add[i]=0;
}
void update(int i,int l,int r,int q1,int q2,int num)
{if(q1<=l&&r<=q2){add[i]=num;sum[i]=(r-l+1)*num;return;}int m=(l+r)/2;if(add[i])pushdown(i,r-l+1);if(q1<=m)update(lson,q1,q2,num);if(q2>m)update(rson,q1,q2,num);sum[i]=sum[i<<1]+sum[i<<1|1];
}
int query(int i,int l,int r,int q1,int q2)
{if(q1<=l&&r<=q2) return sum[i];if(add[i])pushdown(i,r-l+1);int m=(l+r)/2;int t1=0,t2=0;if(q1<=m)t1=query(lson,q1,q2);if(q2>m)t2=query(rson,q1,q2);return t1+t2;
}
int main()
{
// freopen("1.in","r",stdin);int l,r,c,num;scanf("%d",&n);{build(1,1,n);scanf("%d",&m);for(int i=1;i<=m;i++){scanf("%d",&c);if(c==0){scanf("%d%d",&l,&r);printf("%d\n",query(1,1,n,l,r));}else{scanf("%d%d%d",&l,&r,&num);update(1,1,n,l,r,num);}}}return 0;
}
1080 更为复杂的买卖房屋姿势
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define eps 10^(-6)
#define Q_CIN ios::sync_with_stdio(false);
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);
#define MOD 10009
#define debug(x) cout<<#x<<":"<<(x)<<endl;
#define lson i<<1,l,m
#define rson i<<1|1,m+1,r
int sum[100010<<2],add[100010<<2],eva[100010<<2];int n,m;
void build(int i,int l,int r)
{add[i]=0,eva[i]=0;if(l==r){scanf("%d",&sum[i]);return;}int m=(l+r)/2;build(lson);build(rson);sum[i]=sum[i<<1]+sum[i<<1|1];
}
void pushdown_eva(int i,int m)
{eva[i<<1]=eva[i<<1|1]=eva[i];sum[i<<1]=eva[i]*(m-m/2);sum[i<<1|1]=eva[i]*(m/2);eva[i]=0;add[i<<1]=add[i<<1|1]=0; //赋值标记使得子区间的加值标记无效
}
void pushdown_add(int i,int m)
{add[i<<1]+=add[i];add[i<<1|1]+=add[i];sum[i<<1]+=add[i]*(m-m/2);sum[i<<1|1]+=add[i]*(m/2);add[i]=0;
}
void update(int i,int l,int r,int q1,int q2,int num,int c)
{if(q1<=l&&r<=q2){if(c){sum[i]=(r-l+1)*num;eva[i]=num;add[i]=0; //add标记被覆盖}else{sum[i]+=(r-l+1)*num;add[i]+=num;}return;}int m=(l+r)/2;if(eva[i])pushdown_eva(i,r-l+1);if(add[i])pushdown_add(i,r-l+1);if(q1<=m)update(lson,q1,q2,num,c);if(q2>m)update(rson,q1,q2,num,c);sum[i]=sum[i<<1]+sum[i<<1|1];
}
int query(int i,int l,int r,int q1,int q2)
{if(q1<=l&&r<=q2) return sum[i];if(eva[i])pushdown_eva(i,r-l+1);if(add[i])pushdown_add(i,r-l+1);int m=(l+r)/2;int t1=0,t2=0;if(q1<=m)t1=query(lson,q1,q2);if(q2>m)t2=query(rson,q1,q2);return t1+t2;
}
int main()
{
// freopen("1.in","r",stdin);int l,r,c,num;scanf("%d%d",&n,&m);{n++;build(1,1,n);for(int i=1;i<=m;i++){scanf("%d%d%d%d",&c,&l,&r,&num);l++;r++;update(1,1,n,l,r,num,c);printf("%d\n",query(1,1,n,1,n));}}return 0;
}
1044 状态压缩 一
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define eps 10^(-6)
#define Q_CIN ios::sync_with_stdio(false);
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);
#define MOD 10009
#define debug(x) cout<<#x<<":"<<(x)<<endl;
#define lson i<<1,l,m
#define rson i<<1|1,m+1,r
int dp[1005][1030];
int sum(int n)
{int ans=0;while(n){ans+=n&1;n>>=1;}return ans;
}
int main()
{
// freopen("1.in","r",stdin);int n,m,q;int a[1005];while(~scanf("%d%d%d",&n,&m,&q)){for(int i=1;i<=n;i++)scanf("%d",&a[i]);
// memset(dp,0,sizeof(dp));for(int i=1;i<=n;i++)for(int j=0;j<(1<<m)-1;j++){if(sum(j)<q)dp[i][j]=max(dp[i-1][(j/2)+(1<<(m-2))],dp[i-1][j/2]);else if(sum(j)>q)dp[i][j]=0;elsedp[i][j]=dp[i-1][j/2];if((j>>(m-2))&1) dp[i][j]+=a[i];}int ans=0;for(int j=0;j<(1<<m)-1;j++)ans=max(dp[n][j],ans);printf("%d\n",ans);}return 0;
}
1089 最短路径 二：Floyd算法
#include <cstdio>
#include <string.h>
#include <vector>
#include <algorithm>
#include <iostream>
#define rep( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define clr( a , x ) memset ( a , x , sizeof (a) );
using namespace std;
const int maxn = 105;
const int inf = 0x3f3f3f3f;int n,m;
int dp[maxn][maxn];void floyd(){rep(k,1,n){rep(i,1,n){rep(j,1,n){dp[i][j] = min(dp[i][j],dp[i][k]+dp[k][j]);}}}
}int main(){int a,b,c;while(cin>>n>>m){clr(dp,inf);rep(i,1,n)dp[i][i]=0;rep(i,1,m){cin>>a>>b>>c;dp[a][b] = dp[b][a] = min(dp[a][b],c);}floyd();rep(i,1,n){rep(j,1,n-1){cout<<dp[i][j]<<" ";}cout<<dp[i][n]<<endl;}}return 0;
}1097 最小生成树一 Prim算法
#include <iostream>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <cmath>using namespace std;
const int maxn = 1e3+50;
const int maxM = 25300*502;
int dis[maxn];
int vis[maxn];
int n;
int mp[maxn][maxn];int prim(int src){for (int i = 1; i <= n; ++i){dis[i] = mp[src][i];}memset(vis,0,sizeof(vis));dis[src] = 0;vis[src] = 1;int ret = 0;for (int i = 1; i < n; ++i){int minDis = 9999999,minIdx;for (int j = 1; j <= n; ++j){if(!vis[j] && dis[j] < minDis){minDis = dis[j];minIdx = j;}}vis[minIdx] = 1;ret += minDis;for (int j = 1; j <= n; ++j){// if(!vis[j] && dis[minIdx] + mp[minIdx][j] < dis[j]){// dis[j] = dis[minIdx] + mp[minIdx][j];// }if(!vis[j] && mp[minIdx][j] < dis[j]){dis[j] = mp[minIdx][j];}}}return ret;
}int main() {while (cin >> n ) {for (int i = 1; i <= n; ++i) {for (int j = 1; j <= n; ++j) {cin >> mp[i][j];}}memset(dis, 0, sizeof(dis));int ret = prim(1);cout << ret << endl;}return 0;
}
1142 三分·三分求极值
#include <cstdio>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
double a,b,c,x,y;double val(double X){return sqrt((X-x)*(X-x)+(a*X*X+b*X+c-y)*(a*X*X+b*X+c-y));
}double solve(double l,double r){double eps = 1e-5;while(l+eps<r){double lmid = l + (r-l)/3,rmid = r - (r-l)/3;if(val(lmid) < val(rmid)){r = rmid;}else{l = lmid;}}return val(l);
}
int main(){cin>>a>>b>>c>>x>>y;printf("%.3f\n", solve(-200.0,200.0));return 0;
}
1183 连通性一·割边与割点
Tarjan算法简化版，注意是无向边 和 边上点的大小关系
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll long long
#define clr( a , x ) memset ( a , x , sizeof (a) );
#define RE freopen("1.in","r",stdin);
#define WE freopen("1.out","w",stdout);
#define SpeedUp std::cout.sync_with_stdio(false);
#define debug(x) cout << "Line " << __LINE__ << ": " << #x << " = " << x << endl;
const int maxn = 20005;
const int maxm = 200052; //~无向边
const int inf = 0x3f3f3f3f;
int head[maxn];
int dfn[maxn], low[maxn];
int cnt;
bool vis[maxn];
int next[maxn];
int father[maxn];
int eCnt;
struct Edge
{int v, next;
} edge[maxm];
void add(int u, int v) {edge[eCnt].v = v, edge[eCnt].next = head[u], head[u] = eCnt++;edge[eCnt].v = u, edge[eCnt].next = head[v], head[v] = eCnt++;
}
struct B
{int u,v;bool operator < (const struct B bb) const{if(u==bb.u)return v<bb.v;return u<bb.u;}
}bridge[maxm]; //存答案的边
int bCnt;
bool node[maxn];
void dfs(int u) {dfn[u] = low[u] = cnt++;vis[u] = true;int child = 0;for (int i = head[u]; i != -1; i = edge[i].next) {int v = edge[i].v;if (!vis[v]) {child++;father[v] = u;dfs(v);low[u] = min(low[u], low[v]);if (father[u] == -1 && child >= 2) {node[u] = true;}if (father[u] != -1 && low[v] >= dfn[u]) {node[u] = true;}if (low[v] > dfn[u]) {bridge[bCnt].u = min(u,v);bridge[bCnt++].v = max(u,v);}}else if (v != father[u]) {low[u] = min(low[u], dfn[v]);}}
}
int main() {// REint n, m;int a, b;while (cin >> n >> m) {eCnt = 0;clr(head, -1);clr(vis, false);clr(father,-1);clr(node,false);clr(dfn,0);clr(low,0);cnt = 0;bCnt = 0;for (int i = 0; i < m; ++i) {cin >> a >> b;add(a, b);}dfs(1);int first = true;int no = 1;for (int i = 1; i <= n; ++i){if(node[i]){if(first){first = false;cout<<i;}else{cout<<" "<<i;}no = 0;}}if(no){cout<<"Null";}cout<<endl;sort(bridge,bridge+bCnt);for (int i = 0; i < bCnt; ++i){cout<<bridge[i].u<<" "<<bridge[i].v<<endl;}// for (int i = 1; i <= n; ++i){// cout<<"dfn="<<dfn[i]<<" low="<<low[i]<<endl;// }}return 0;
}1066 无间道之并查集
https://hihocoder.com/problemset/solution/1271732
#include <iostream>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <cmath>
#include <map>using namespace std;
const int maxn = 100005;
const int maxM = 25300*502;int root[maxn*2];
int n;
int findRoot(int cur){if(cur == root[cur]){return cur;}return root[cur] = findRoot(root[cur] );
}int merge(int x,int y){int rootX = findRoot(x);int rootY = findRoot(y);if(rootX != rootY){root[rootX] = rootY;}
}int main() {int op;string str1,str2;map<string,int>mp;while (cin >> n ) {for (int i = 0; i <= n; i++) {root[i] = i;}mp.clear();int cnt = 0;for (int i = 0; i < n; i++) {cin >> op >> str1 >> str2;int idx1 ,idx2 ;idx1 = mp[str1];idx2 = mp[str2];if( mp[str1] == 0){idx1 = mp[str1] = ++cnt;}if( mp[str2] == 0){idx2 = mp[str2] = ++cnt;} if(op==1){// cout<< idx1 << " " << idx2 << " === "<< endl;if( findRoot(idx1) == findRoot(idx2) ){cout<<"yes"<<endl;}else{cout<<"no"<<endl;}}else{merge(idx1,idx2);// for (int i = 0; i < cnt; i++) {// cout<<word[i]<< " " << i << " "<<findRoot(i) << endl;// }}}}return 0;
}
1038 01背包
https://hihocoder.com/problemset/solution/1270589
#include <iostream>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <cmath>using namespace std;
const int maxN = 555;
const int maxM = 25300*502;int main() {int n, m;int cost[maxN], val[maxN];int dp[maxM];while (cin >> n >> m) {for (int i = 0; i < n; ++i) {cin >> cost[i] >> val[i];}memset(dp, 0, sizeof(dp));int ans = 0;for (int i = 0; i < n; ++i) {for (int j = m; j >= cost[i]; --j) {dp[j] = max(dp[j], dp[j - cost[i]] + val[i]);}}cout << dp[m] << endl;}return 0;
}
1081 最短路径·一
https://hihocoder.com/problemset/solution/1271836 注意路是双向的，跟prim就差几行代码
#include <iostream>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <cmath>using namespace std;
const int maxn = 1e3+50;
const int maxM = 25300*502;
const int inf = 0x3f3f3f3f;
int dis[maxn];
int vis[maxn];
int n;
int mp[maxn][maxn];int dijkstra(int src,int dst){for (int i = 1; i <= n; ++i){dis[i] = mp[src][i];}memset(vis,0,sizeof(vis));dis[src] = 0; vis[src] = 1;for (int i = 1; i < n; ++i){int minDis = inf,minIdx = 0;for (int v = 1; v <= n; ++v){if(!vis[v] && dis[v] < minDis){minDis = dis[v];minIdx = v;}}vis[minIdx] = 1;for (int v = 1; v <= n; ++v){if(!vis[v] && dis[minIdx] + mp[minIdx][v] < dis[v]){dis[v] = dis[minIdx] + mp[minIdx][v];}}}return dis[dst];
}int main() {int s,t,a,b,val,m;while (cin >> n >> m >> s >> t) {memset(mp, inf, sizeof(mp));for (int i = 0; i < m; ++i) {cin >> a >> b >> val;mp[b][a] = mp[a][b] = min(mp[a][b],val);}int ret = dijkstra(s,t);cout << ret << endl;}return 0;
}
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