CodeVs天梯钻石Diamond题解

title: CodeVs天梯之Diamond
date: 2017-12-28
tags:

天梯
CodesVs
categories: OI

CodeVs刷题攻略之Diamond

2018.1.14 By gwj1139177410

0x01最短路

Car的旅行路线

//1.计算几何求第四点坐标, 方法很多
//2.虚点，到A城市的四个机场边权都为0
//3.SPFA跑最短路
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<vector>
#include<queue>
using namespace std;//Timu
double n, m, a, b;
double x[500], y[500], t[110];//Graph
struct Edge{int to;  double w;Edge(int x, double y):to(x),w(y){}
};
vector<Edge>G[500];
void insert(int u,int v,double w){G[u].push_back(Edge(v,w));
}
double distant(double nx,double ny,double mx,double my){return sqrt((nx-mx)*(nx-mx)+(ny-my)*(ny-my));
}//spfa
queue<int>q;
int vis[500];
double dis[500];
void spfa(){while(!q.empty()){int u = q.front();  q.pop();  vis[u] = 0;for(int i = 0; i < G[u].size(); i++){int v = G[u][i].to;if(dis[u]+G[u][i].w<dis[v]){dis[v] = dis[u]+G[u][i].w;if(!vis[v]){q.push(v);vis[v] = 1;}}}}
}int main(){int T;  cin>>T;while(T--){//1.初始化memset(x,0,sizeof x);memset(y,0,sizeof y);memset(t,0,sizeof t);memset(vis,0,sizeof vis);//2.dateincin>>n>>m>>a>>b;for(int i = 0; i < n; i++){for(int j = 1; j < 4; j++)cin>>x[i*4+j]>>y[i*4+j];cin>>t[i+1];//point4, double l1=distant(x[i*4+1],y[i*4+1],x[i*4+2],y[i*4+2]);double l2=distant(x[i*4+1],y[i*4+1],x[i*4+3],y[i*4+3]);double l3=distant(x[i*4+2],y[i*4+2],x[i*4+3],y[i*4+3]);double l=max(l1,max(l2,l3));//三条里最常的就是对角线,然后中点坐标得到第4点if(l1 == l)x[i*4+4]=(x[i*4+1]+x[i*4+2])-x[i*4+3], y[i*4+4]=(y[i*4+1]+y[i*4+2])-y[i*4+3];if(l2 == l)x[i*4+4]=(x[i*4+1]+x[i*4+3])-x[i*4+2], y[i*4+4]=(y[i*4+1]+y[i*4+3])-y[i*4+2];if(l3 == l)x[i*4+4]=(x[i*4+2]+x[i*4+3])-x[i*4+1], y[i*4+4]=(y[i*4+2]+y[i*4+3])-y[i*4+1];}//3.预处理,建图(把所有机场连起来就好啦啦啦~)n *= 4;for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){if(i == j)continue;int c1 = (i-1)/4+1, c2 = (j-1)/4+1;if(c1 == c2){double w = distant(x[i],y[i],x[j],y[j])*t[c1];insert(i,j,w);insert(j,i,w);}else{double w = distant(x[i],y[i],x[j],y[j])*m;insert(i,j,w);insert(j,i,w);}}}//4.runfor(int i = 1; i <= n; i++)dis[i] = 1e9+1;for(int i = (a-1)*4+1; i <= a*4; i++){q.push(i);dis[i] = 0;vis[i] = 1;}spfa();//5.dateoutdouble ans = 1e9;for(int i = (b-1)*4+1; i <= 4*b; i++)ans = min(ans, dis[i]);printf("%.1lf\n",ans);}return 0;
}

多源最短路

//Floyd-wallshall模板
#include<iostream>
using namespace std;
int n, e[110][110];
int main(){ios::sync_with_stdio(false);cin>>n;for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)cin>>e[i][j];for(int k = 1; k <= n; k++)for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)   if(e[i][k]+e[k][j]<e[i][j])e[i][j] = e[i][k]+e[k][j];int T;  cin>>T;while(T--){int a, b;cin>>a>>b;cout<<e[a][b]<<"\n";}return 0;
}

回家

//数据太小, Floyd一番水
#include<iostream>
#include<algorithm>
using namespace std;
int n = 60, m;
int e[1010][1010], vis[1010];
int main(){cin>>m;for(int i = 0; i < n; i++)for(int j = 0; j < n; j++)e[i][j] = i==j ? 0 : 0xffffff;for(int i = 0; i < m; i++){char a, b; int w;  cin>>a>>b>>w;int x = a-'A', y = b-'A';if(x>=0 && x<25)vis[x] = 1;if(y>=0 && y<25)vis[y] = 1;//bugs数据可能有覆盖e[x][y] = min(e[x][y], w);e[y][x] = min(e[y][x], w);}for(int k = 0; k < n; k++)for(int i = 0; i < n; i++)for(int j = 0; j < n; j++)e[i][j] = min(e[i][j], e[i][k]+e[j][k]);int ans2 = 0xffffff, ans1;for(int i = 0; i < n; i++)if(vis[i] && e[i][(int)'Z'-'A']<ans2){ans2 = e[i][(int)'Z'-'A'];ans1 = i;}cout<<(char)(ans1+'A')<<" "<<ans2;return 0;
}

//标程Dijkstra
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;//T
int n = 60, m, vis[1010];//Graph
struct Edge{int v, w;Edge(int v, int w):v(v),w(w){}
};
vector<Edge>G[70];//Dijkstra
int dis[70], book[70], s = (int)'Z'-'A';
void Dijkstra(){for(int i = 0; i < n; i++)dis[i] = 0xffffff;for(int i = 0; i < G[s].size(); i++)//bugs数据可能有覆盖dis[G[s][i].v] = min(dis[G[s][i].v],G[s][i].w);dis[s] = 0;  book[s] = 1;for(int i = 0; i < n; i++){int v, w=0xffffff;for(int j = 0; j < n; j++)if(!book[j] && dis[j]<w)w = dis[v=j];book[v] = 1;for(int j = 0; j < G[v].size(); j++)dis[G[v][j].v] = min(dis[G[v][j].v], w+G[v][j].w);}
}int main(){cin>>m;for(int i = 1; i <= m; i++){char a, b; int w;  cin>>a>>b>>w;int x = a-'A', y = b-'A';if(x>=0 && x<25)vis[x] = 1;if(y>=0 && y<25)vis[y] = 1;G[x].push_back(Edge(y,w));G[y].push_back(Edge(x,w));}Dijkstra();int v, w=0xffffff;for(int i = 0; i < n; i++)if(vis[i] && dis[i]<w)w = dis[v=i];cout<<(char)(v+'A')<<" "<<w<<"\n";return 0;
}

0x02启发式搜索

靶形数独

//(如果你玩数独会怎么填呢)......启发式:把能确定的(剩余少的)先填上
#include<iostream>
using namespace std;const int score[10][10]={{0,0,0,0,0,0,0,0,0,0},{0,6,6,6,6,6,6,6,6,6},{0,6,7,7,7,7,7,7,7,6},{0,6,7,8,8,8,8,8,7,6},{0,6,7,8,9,9,9,8,7,6},{0,6,7,8,9,10,9,8,7,6},{0,6,7,8,9,9,9,8,7,6},{0,6,7,8,8,8,8,8,7,6},{0,6,7,7,7,7,7,7,7,6},{0,6,6,6,6,6,6,6,6,6}
};//r[i][j],第i行第j个数是否填过...row_cnt[i],第i行填了几个数
int a[11][11], row[11][11], col[11][11], area[11][11];
int row_cnt[11], col_cnt[11], cnt, ans=-1;
//得到(r,c)是第几个区域的
inline int id(int r, int c){ return (r-1)/3*3+(c-1)/3+1; }
//计算当前得分
inline int calc(){int sum = 0;for(int i = 1; i <= 9; i++)for(int j = 1; j <= 9; j++)sum += score[i][j]*a[i][j];return sum;
}void dfs(int r, int c, int cc){if(cc == 81){ans = max(ans, calc());return ;}else for(int i = 1; i <= 9; i++){//尝试每个填数if(row[r][i]||col[c][i]||area[id(r,c)][i]) continue;row[r][i] = col[c][i] = area[id(r,c)][i] = 1;row_cnt[r]++;  col_cnt[c]++;a[r][c] = i;//找没有填的最少的行和列int tr, vr=-1, tc, vc=-1;for(int j = 1; j <= 9; j++)if(row_cnt[j]>vr && row_cnt[j]!=9)vr = row_cnt[tr=j];for(int j = 1; j <= 9; j++)if(col_cnt[j]>vc && !a[tr][j])//(r,c)未填数vc = col_cnt[tc=j];dfs(tr,tc,cc+1);row[r][i] = col[c][i] = area[id(r,c)][i] = 0;row_cnt[r]--;  col_cnt[c]--;a[r][c] = 0;}
}int main(){//dateinfor(int i = 1; i <= 9; i++){for(int j = 1; j <= 9; j++){cin>>a[i][j];if(a[i][j]){//更新初始数据row[i][a[i][j]] = col[j][a[i][j]] = area[id(i,j)][a[i][j]] = 1;row_cnt[i]++;  col_cnt[j]++;  cnt++;}}}//找没有填的最少的行和列。int tr, vr=-1, tc, vc=-1;for(int i = 1; i <= 9; i++)if(row_cnt[i]>vr && row_cnt[i]!=9)vr = row_cnt[tr=i];for(int i = 1; i <= 9; i++)if(col_cnt[i]>vc && !a[tr][i])vc = col_cnt[tc=i];//dfsdfs(tr,tc,cnt);cout<<ans<<"\n";return 0;
}

八数码难题

#include<iostream>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
using namespace std;
const int dx[] = {0,0,-1,1};
const int dy[] = {1,-1,0,0};
string goal = "123804765", s;
map<string,int>ma;
int main(){cin>>s;queue<string>q;q.push(s);while(!q.empty()){string t = q.front();  q.pop();if(t == goal)break;int z;for(z = 0; z < 9; z++)if(t[z]=='0')break;int x=z/3, y=z%3;for(int i = 0; i < 4; i++){int nx=x+dx[i], ny=y+dy[i], nz=nx*3+ny;if(nx<0||nx>=3||ny<0||ny>=3)continue;string tt = t;swap(tt[z],tt[nz]);if(!ma.count(tt)){q.push(tt);ma[tt] = ma[t]+1;}}}cout<<ma[goal]<<"\n";return 0;
}

0x03线段树入门

线段树练习

#include<iostream>
using namespace std;
const int maxn = 100010;
#define lch p<<1
#define rch p<<1|1
struct node{int val, addmark;
}sgt[maxn<<2];
void pushdown(int p, int l, int r){if(!sgt[p].addmark)return;int t = sgt[p].addmark, m=l+r>>1;sgt[lch].addmark += t;sgt[rch].addmark += t;sgt[lch].val += t*(m-l+1);sgt[rch].val += t*(r-m);sgt[p].addmark = 0;
}
void update(int p, int l, int r, int ql, int qr, int v){if(l>qr || r<ql)return ;if(ql<=l && r<=qr){sgt[p].val += v*(r-l+1);sgt[p].addmark += v;return ;}int m = l+r>>1;if(ql<=m)update(lch,l,m,ql,qr,v);if(qr>m)update(rch,m+1,r,ql,qr,v);sgt[p].val = sgt[lch].val+sgt[rch].val;
}
int query(int p, int l, int r, int ql, int qr){if(ql<=l && r<=qr)return sgt[p].val;pushdown(p,l,r);int m = l+r>>1, ans = 0;if(ql<=m)ans += query(lch,l,m,ql,qr);if(qr>m)ans += query(rch,m+1,r,ql,qr);return ans;
}
int main(){int n, m;cin>>n;for(int i = 1; i <= n; i++){int x;  cin>>x;  update(1,1,n,i,i,x);}cin>>m;for(int i = 1; i <= m; i++){int op, x, y;cin>>op>>x>>y;if(op == 1){update(1,1,n,x,x,y);}else{cout<<query(1,1,n,x,y)<<"\n";}}return 0;
}

线段树练习 2

#include<iostream>
using namespace std;
const int maxn = 100010;
#define lch p<<1
#define rch p<<1|1
struct node{int val, addmark;
}sgt[maxn<<2];
void pushdown(int p, int l, int r){if(!sgt[p].addmark)return;int t = sgt[p].addmark, m=l+r>>1;sgt[lch].addmark += t;sgt[rch].addmark += t;sgt[lch].val += t*(m-l+1);sgt[rch].val += t*(r-m);sgt[p].addmark = 0;
}
void update(int p, int l, int r, int ql, int qr, int v){if(l>qr || r<ql)return ;if(ql<=l && r<=qr){sgt[p].val += v*(r-l+1);sgt[p].addmark += v;return ;}int m = l+r>>1;if(ql<=m)update(lch,l,m,ql,qr,v);if(qr>m)update(rch,m+1,r,ql,qr,v);sgt[p].val = sgt[lch].val+sgt[rch].val;
}
int query(int p, int l, int r, int ql, int qr){if(ql<=l && r<=qr)return sgt[p].val;pushdown(p,l,r);int m = l+r>>1, ans = 0;if(ql<=m)ans += query(lch,l,m,ql,qr);if(qr>m)ans += query(rch,m+1,r,ql,qr);return ans;
}
int main(){int n, m;cin>>n;for(int i = 1; i <= n; i++){int x;  cin>>x;  update(1,1,n,i,i,x);}cin>>m;for(int i = 1; i <= m; i++){int op, x, y, k;cin>>op;if(op == 1){cin>>x>>y>>k;update(1,1,n,x,y,k);}else{cin>>x;cout<<query(1,1,n,x,x)<<"\n";}}return 0;
}

约瑟夫问题

//好玩么
#include<iostream>
using namespace std;
int n, m, a[30010], r, p;
int main(){cin>>n>>m;for(int i = 1; i <= n; i++)a[i]=i;r = n;  p = 1;while(r>1){p = (p+m-1)%r;if(p==0)p=r;cout<<a[p]<<" ";for(int i = p; i <= r-1; i++)a[i]=a[i+1];r--;}cout<<a[1]<<"\n";return 0;
}

0x04并查集

舒适的路线

//类MST,对于每条边把他作为_max且作为常数，然后枚举所有比他小的边直到联通，维护全局最大值
#include<iostream>
#include<algorithm>
using namespace std;
const int inf = 0xfffffff;
int n, m, st, ed, _max=inf, _min=1;
int gcd(int a, int b){ return !b?a:gcd(b,a%b);}
//graph
struct edge{ int u, v, w;}e[5010];
bool cmp(edge a, edge b){ return a.w<b.w; }
//UnionFindSet
int fa[510];
void init(int n){ for(int i = 1; i <= n; i++) fa[i]=i; }
int find(int x){ return x==fa[x]?x:fa[x]=find(fa[x]);}
void merge(int x, int y){ if(find(x)==find(y))return ; fa[find(x)]=find(y); }
//main
int main(){cin>>n>>m;for(int i = 1; i <= m; i++)cin>>e[i].u>>e[i].v>>e[i].w;cin>>st>>ed;sort(e+1,e+m+1, cmp);for(int i = 1; i <= m; i++){init(n);for(int j = i; j >= 1; j--){merge(e[j].u, e[j].v);if(find(st)==find(ed)){if((e[i].w*1.0/e[j].w) < (_max*1.0/_min)){_max = e[i].w;_min = e[j].w;}break;}}}if(_max==inf && _min==1){ cout<<"IMPOSSIBLE\n"; return 0;}int r = gcd(_max, _min);_max /= r;_min /= r;if(_min == 1)cout<<_max<<"\n";else cout<<_max<<"/"<<_min<<"\n";return 0;
}

关押罪犯

//并查集及补集
//凡是与i+n节点在同一个集合里的，都是不能与i在同一个集合里的。
#include<iostream>
#include<algorithm>
using namespace std;
struct Edge{ int u, v, w; }e[100010];
bool cmp(Edge a, Edge b){ return a.w>b.w; }
int fa[20010<<1];
void init(int n){ for(int i = 1; i <= n; i++)fa[i]=i; }
int find(int x){ return x==fa[x]?x:fa[x]=find(fa[x]); }
void merge(int x, int y){ x=find(x);y=find(y);if(x!=y)fa[x]=y;}
int main(){int n, m;cin>>n>>m;for(int i = 1; i <= m; i++)cin>>e[i].u>>e[i].v>>e[i].w;sort(e+1,e+m+1,cmp);init(n<<1);for(int i = 1; i <= m; i++){int u = e[i].u, v = e[i].v;if(find(u) == find(v)){cout<<e[i].w<<"\n";  return 0;}merge(u+n, v);merge(v+n, u);}cout<<0<<"\n";//....return 0;
}

家族

//并查集模板
#include<iostream>
using namespace std;
int fa[5010];
void init(int n){for(int i = 1; i <= n; i++)fa[i]=i;}
int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
void merge(int x, int y){x=find(x);y=find(y);if(x!=y)fa[x]=y;}
int main(){int n, m, p;cin>>n>>m>>p;init(n);for(int i = 1; i <= m; i++){int a, b;  cin>>a>>b;  merge(a,b);}for(int i = 1; i <= p; i++){int a, b;  cin>>a>>b;if(find(a)==find(b))cout<<"Yes\n";else cout<<"No\n";}return 0;
}

食物链

//本题思路：把x作为a,b,c三种动物分别加入，维护三个集合的关系。
//并查集及补集
//其中i用来连接与i同类的，i+n用来连接能吃i的，i+2*n用来连接i能吃的。
//具体来说，凡是与i+n节点在同一个集合里的，都是被i吃的动物。
#include<iostream>
#include<algorithm>
using namespace std;
int ans = 0;
int fa[50010*3];
void init(int n){ for(int i = 1; i <= n; i++)fa[i]=i; }
int find(int x){ return x==fa[x]?x:fa[x]=find(fa[x]); }
void merge(int x, int y){ x=find(x);y=find(y);if(x!=y)fa[x]=y;}
int some(int x, int y){ return find(x)==find(y); }
int main(){int n, m;cin>>n>>m;init(n*3);for(int i = 1; i <= m; i++){int op, x, y;cin>>op>>x>>y;if(x>n||y>n){ans++;continue;}//2)当前的话中X或Y比N大，就是假话if(op==2&&x==y){ans++;continue;}//3)当前的话表示X吃X，就是假话if(op==1){if(some(x,y+n)||some(y,x+n))ans++;//如果x吃y或者y吃x，就不是同类else{merge(x,y);//x和y是同类merge(x+n,y+n);//能吃x和y的也是同类merge(x+n*2,y+n*2);//x和y能吃的也是同类}}else{if(some(x,y)||some(y,x+n))ans++;//如果x和y是同类或者y吃xelse{merge(x,y+n);//x和吃y的连起来merge(x+n,y+n*2);//能吃x的和被y吃的连起来(三种动物之间的关系啊)merge(x+n*2,y);//x能吃的和y连起来}}}cout<<ans<<"\n";return 0;
}

0x05堆

地鼠游戏

//贪心：优先打价值最大的(如果能打的话)
#include<iostream>
#include<algorithm>
using namespace std;
struct d{ int t, w; }a[110];
bool cmp(d a, d b){ return a.w>=b.w; }
int n, book[110], ans;
int main(){cin>>n;for(int i = 0; i < n; i++)cin>>a[i].t;for(int i = 0; i < n; i++)cin>>a[i].w;sort(a, a+n, cmp);for(int i = 0; i < n; i++)for(int j = a[i].t; j >= 1; j--)if(!book[j]){ ans += a[i].w; book[j] = 1; break; }cout<<ans;return 0;
}

合并果子

//STL默认大根堆
#include<iostream>
#include<queue>
using namespace std;
int n, ans;
priority_queue<int,vector<int>,greater<int> >q;
int main(){cin>>n;for(int i = 1; i <= n; i++){int x;  cin>>x;  q.push(x);}while(q.size()!=1){int a = q.top();  q.pop();int b = q.top();  q.pop();ans += a+b;q.push(a+b);}cout<<ans<<"\n";return 0;
}

最小的N个和

//动态维护大根堆,贪心减少入队元素个数
//以及,拓展参见刘汝佳蓝书P189,K路归并
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 100010;
int n, a[maxn], b[maxn];
//q为答案的n个元素。
priority_queue<int>q;
//递归输出
void print(){if(q.empty())return ;int t = q.top();  q.pop();print();cout<<t<<" ";
}
int main(){cin>>n;for(int i = 1; i <= n; i++)cin>>a[i];for(int i = 1; i <= n; i++)cin>>b[i];//step1:排序,让序列单调，后面用单调性减少状态数sort(a+1,a+n+1);sort(b+1,b+n+1);//step2:随便加n个元素作为初始值for(int i = 1; i <= n; i++){q.push(a[1]+b[i]);}for(int i = 2; i <= n; i++){if(a[i]+b[1]>=q.top())break;//step3因为单调,所以后面的a[i]+b[1]只会更大。for(int j = 1; j <= n; j++){if(a[i]+b[j]>=q.top())break;//step4:因为单调,所以后面的肯定会更大。//step5:如果没有break,则当前元素比答案中的最大值要大，更新答案。q.pop();q.push(a[i]+b[j]);}}//step6:此时队列中剩下的n各元素就是最小值print();return 0;
}

0x06高精度++

高精度练习之除法

//高精除高精,模板
//思路:模拟减法，a每次减去b的10^n倍可以提高效率
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
const int maxn = 510;
int a[maxn], b[maxn], c[maxn], t[maxn];
//比较x和y的大小,x>y时返回1,x==y返回0,x<y返回-1
int compare(int x[], int y[]){if(x[0] < y[0])return -1;if(x[0] > y[0])return 1;for(int i = x[0]; i > 0; i--){ //越后面的位越大啊if(x[i] < y[i])return -1;if(x[i] > y[i])return 1;}return 0;
}
//y=x*(10^k)
void times(int x[], int y[], int k){for(int i = 1; i <= x[0]; i++)y[i+k-1] = x[i];y[0] = x[0]+k-1;
}
int main(){string s1, s2;  cin>>s1>>s2;a[0] = s1.size();  b[0] = s2.size();//bugs:a比b小的情况,就是...如果只剩0的话这个商的长度的也会变成0 错误数据:0 100if(a[0]<b[0]||(a[0]==b[0]&&s1<s2)){ cout<<"0\n"; return 0; }for(int i = 1; i <= a[0]; i++)a[i] = s1[a[0]-i]-'0';for(int i = 1; i <= b[0]; i++)b[i] = s2[b[0]-i]-'0';c[0] = a[0]-b[0]+1;for(int i = c[0]; i > 0; i--){memset(t,0,sizeof(t));times(b,t,i);while(compare(a,t)>=0){c[i]++;//高精减,a=a-tfor(int j = 1; j <= a[0]; j++){if(a[j] < t[j]){a[j+1]--;a[j] += 10;}a[j] -= t[j];}while(!a[a[0]] && a[0]>1)a[0]--;}}while(!c[c[0]] && c[0]>1)c[0]--;for(int i = c[0]; i > 0; i--)cout<<c[i];return 0;
}

高精度练习之大整数开根

//二分答案
#include<iostream>
#include<string>
#include<cstring>
using namespace std;
const int maxn = 1010;
int a[maxn], l[maxn], r[maxn], m[maxn], t[maxn];
//m=l+r>>1
void mid(){//m = l;memcpy(m,l,sizeof(l));//m += r;m[0] = r[0]+1;for(int i = 1; i <= r[0]; i++){m[i] += r[i];if(m[i]>=10){m[i] %= 10;m[i+1]++;}}while(m[0]>1 && m[m[0]]==0)m[0]--;//m /= 2;for(int i = m[0]; i >= 1; i--){if(i > 1)m[i-1]+=m[i]%2*10;m[i] /= 2;}while(m[0]>1 && m[m[0]]==0)m[0]--;
}
//return m*m>a;
bool C(){//t = 0;memset(t,0,sizeof(t));//t = m*m;for(int i = 1; i <= m[0]; i++)for(int j = 1; j <= m[0]; j++)t[i+j-1] += m[i]*m[j];t[0] = m[0]*2;for(int i = 1; i <= t[0]; i++){//处理进位t[i+1] += t[i]/10;t[i] %= 10;}while(t[0]>1 && t[t[0]]==0)t[0]--;//return t>a;if(t[0] != a[0])return t[0]>a[0];for(int i = t[0]; i >= 1; i--)if(t[i]!=a[i])return t[i]>a[i];return 0;
}
int main(){//输入string s;  cin>>s;a[0] = s.size();for(int i = 1; i <= a[0]; i++)a[i] = s[a[0]-i]-'0';//二分l[0] = 1;r[0] = a[0]/2+2;r[r[0]] = 1;for(int i = 1; i <= 2000; i++){mid();  //m=l+r>>1;if(C())memcpy(r,m,sizeof(m));//r=m;else memcpy(l,m,sizeof(m));//l=m;}//输出for(int i = l[0]; i >= 1; i--)cout<<l[i];return 0;
}

0x07哈希表

元素查找

#include<iostream>
#include<set>
using namespace std;
set<int>s;
int main(){int n, m;cin>>n>>m;for(int i = 1; i <= n; i++){int x;  cin>>x;  s.insert(x);}for(int i = 1; i <= m; i++){int x;  cin>>x;if(s.count(x))cout<<"YES\n";else cout<<"NO\n";}return 0;
}

互斥的数

/*
贪心
1.找出不互质的数的集合，就是把互斥的数删去.
2.那么当有两个互斥的数时，如果删掉前面(小)的，这个数后面的与它互斥的数也会入选，所以删掉后面的更优。
3.因为每个数都是不同的。
*/
#include<iostream>
#include<algorithm>
#include<map>
using namespace std;
const int maxn = 1e5+10;
int n, p, a[maxn], ans;
map<int,int>ma;
int main(){cin>>n>>p;for(int i = 1; i <= n; i++)cin>>a[i];sort(a+1,a+n+1);//枚举每个数，如果当前数没有被删，那么集合元素+1，然后把与他互斥的数删了。for(int i = 1; i <= n; i++)if(!ma[a[i]]){ ma[a[i]*p]=1; ans++; }cout<<ans<<"\n";return 0;
}

砝码称重 2

//Meet in the Middle
#include<iostream>
#include<algorithm>
#include<map>
using namespace std;
int n, mass, ans(666), f[233];
map<int, int>ma; //能称出的质量->需要的砝码
//k:当前用的砝码个数,cur:从哪个砝码开始选,sum:当前称出的质量
void dfs1(int k, int cur, int sum){if(sum>mass || cur>n/2)return ;ma[sum] = k;dfs1(k+1,cur+1,sum+f[cur]);//选当前砝码dfs1(k,cur+1,sum);//不选当前砝码
}
void dfs2(int k, int cur, int sum){if(sum>mass || cur>n)return ;if(ma.find(mass-sum) != ma.end()){//如果能跟前半段的结果组成目标质量ans = min(ans,k+ma[mass-sum]);//更新答案return;}dfs2(k+1,cur+1,sum+f[cur]);dfs2(k,cur+1,sum);
}
int main(){cin>>n>>mass;for(int i = 0; i < n; i++)cin>>f[i];dfs1(0,0,0);//先搜前半段dfs2(0,n/2,0);//再搜后半段cout<<ans<<"\n";return 0;
}

0x08树型动态规划

访问艺术馆

/*
1.将博物馆的结构抽象成一棵二叉树，每条边都有对应的权值（走过这条边花费的时间）
2.只在叶子节点有藏画，要求你在有限的时间内偷到尽可能多的藏画。
3.点的信息按照深度优先顺序给出（前序遍历），建立一颗二叉树；
4.然后从根节点开始深搜，每走过一条走廊到达下一个点，
5.剩余的时间remain要减去2倍这条走廊的花费，相当于一去一回；
*/
//f[i,j]:来到第i个走廊（还未走过这条走廊）还剩下j时间，能拿到最大的画的数量。
//f[i,j]=max{f[i,j],f[lch,k]+f[rch,j-2*t[i]-k]| 0<=k<=j-2*t[i]};
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1010;
//堆式建树
int tree[maxn<<1][2], f[maxn][maxn];
void init(int root){//tree[i][0]:走过第i条走廊的时间，tree[i][1]:第i条走廊某端的藏画cin>>tree[root][0]>>tree[root][1];tree[root][0] *= 2;if(!tree[root][1]){init(root<<1);init(root<<1|1);}
}
int dp(int i, int j){if(f[i][j] || j==0)return f[i][j];//搜过了或者没有时间了就返回if(tree[i][1])return f[i][j]=min(tree[i][1],(j-tree[i][0])/5);//有藏画的叶子节点for(int k = 0; k <= j-tree[i][0]; k++)f[i][j]=max(f[i][j],dp(i<<1,k)+dp(i<<1|1,j-tree[i][0]-k));return f[i][j];
}
int main(){int tot;cin>>tot;init(1);cout<<dp(1,tot)<<"\n";return 0;
}

没有上司的舞会

//用f[x][0],f[x][1] 分别表示x没去和去了的最大价值。
//f[x][0] = sigmar:max(f[y][0],f[y][1]);
//f[x][1] = sigmar:f[y][0];
#include<iostream>
#include<algorithm>
#include<vector>
const int maxn = 6000<<1;
using namespace std;
int n, r[maxn], f[maxn][2], in[maxn];
vector<int>G[maxn];
int dp(int x, int q){if(f[x][q])return f[x][q];if(q)f[x][q] = r[x];for(int i = 0; i < G[x].size(); i++){int y = G[x][i];if(q)f[x][q] += dp(y,0);else f[x][q] += max(dp(y,0),dp(y,1));}return f[x][q];
}
int main(){cin>>n;for(int i = 1; i <= n; i++)cin>>r[i];for(int i = 1; i <= n; i++){int a, b;  cin>>a>>b;G[b].push_back(a);in[a]++;}int head = 0; //找树根,即入度为0的结点for(int i = 1; i <= n; i++)if(in[i]==0){ head = i; break; }cout<<max(dp(head,1),dp(head,0))<<"\n";return 0;
}

0x09最小生成树

最小生成树

//MST-Prim-贪心
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn = 110;
//Graph
int e[maxn][maxn];
//Prim
int dis[maxn], book[maxn];
//main
int main(){ios::sync_with_stdio(false);int n;  cin>>n;for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)cin>>e[i][j];//将1号顶点加入生成树book[1] = 1;for(int i = 1; i <= n; i++)dis[i]=e[1][i];//将剩余的n-1个点加入生成树for(int i = 2; i <= n; i++){//找到所有点里面到生成树距离最短的int v, w=0xffffff;for(int j = 1; j <= n; j++)if(!book[j] && dis[j]<w)w = dis[v=j];//将该点加入生成树book[v] = 1;//用该点的出边松弛其他非生成树点到生成树的距离for(int j = 1; j <= n; j++)if(!book[j] && dis[j]>e[v][j])dis[j] = e[v][j];}LL ans = 0;for(int i = 1; i <= n; i++)ans += dis[i];cout<<ans<<"\n";return 0;
}

//MST-Prim-贪心-堆优化
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 110;
//Graph
int e[maxn][maxn],ans;
//Prim
struct node{int v, w;node(int v=0, int w=0):v(v),w(w){}bool operator < (node b)const{return w>b.w;}
};
priority_queue<node>q;//保存所有可以抵达生成树的边
int book[maxn];
//main
int main(){ios::sync_with_stdio(false);int n;  cin>>n;for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)cin>>e[i][j];//将1号顶点加入生成树book[1] = 1;for(int i = 1; i <= n; i++)if(e[1][i])q.push(node(i,e[1][i]));//将剩余的n-1个点加入生成树for(int i = 2; i <= n; i++){//找到所有(与生成树相连的)点里面到生成树距离最短的node t = q.top();  q.pop();while(book[t.v]){//只有不在生成树里的点才可以加到生成树里面，这里避免重复。t = q.top();  q.pop();}//将该点加入生成树book[t.v] = 1;  ans += t.w;//用该点的出边松弛其他非生成树点到生成树的距离for(int j = 1; j <= n; j++)if(!book[j] && e[t.v][j])//当前加入生成树的点可以扩充出的边指向的节点q.push(node(j,e[t.v][j]));}cout<<ans<<"\n";return 0;
}

最优布线问题

//MST-Kruskal-排序贪心+并查集
//题中N=M,(M小于N^2的)稀疏图。
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn = 100010;
//Graph
struct Edge{ int u, v, w; }e[maxn];
bool cmp(Edge a, Edge b){return a.w<b.w;}
//UnionFindSet
int fa[maxn];
void init(int n){for(int i=1;i<=n;i++)fa[i]=i;}
int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
void merge(int x,int y){x=find(x);y=find(y);if(x!=y)fa[x]=y;}
//main
int main(){int n, m;cin>>n>>m;for(int i = 1; i <= m; i++)cin>>e[i].u>>e[i].v>>e[i].w;sort(e+1,e+m+1,cmp);LL ans = 0;init(n);for(int i = 1; i <= m; i++){int u = e[i].u, v = e[i].v;if(find(u) != find(v)){merge(u,v);ans += e[i].w;}}cout<<ans<<"\n";return 0;
}

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CodeVs天梯钻石Diamond题解

CodeVs刷题攻略之Diamond

0x01最短路

0x02启发式搜索

0x03线段树入门

0x04并查集

0x05堆

0x06高精度++

0x07哈希表

0x08树型动态规划

0x09最小生成树

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