Codeforces Gym 2015 ACM Arabella Collegiate Programming Contest

比赛链接：

http://codeforces.com/gym/100676

题目链接：

http://codeforces.com/gym/100676/attachments/download/3333/acm-arabella-collegiate-programming-contest-en.pdf

A. Relational Operator

直接模拟，复杂度O(1)。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
int main()
{int T;scanf("%d",&T);int a,b;char c[5];while(T--){scanf("%d%s%d",&a,c,&b);if(strlen(c)==1){if(c[0]=='<')printf("%s\n",(a<b ? "true" : "false"));else printf("%s\n",(a>b ? "true" : "false"));}else{if(c[0]=='!')printf("%s\n",(a!=b ? "true" : "false"));else if(c[0]=='=')printf("%s\n",(a==b ? "true" : "false"));else if(c[0]=='<')printf("%s\n",(a<=b ? "true" : "false"));else printf("%s\n",(a>=b ? "true" : "false"));}}return 0;
}

B. Three Angles

判断三个角度是否均为正且和为180即可，复杂度O(1)。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
int main()
{int T;scanf("%d",&T);int a,b,c;while(T--){scanf("%d%d%d",&a,&b,&c);printf("%s\n",(a>0 && b>0 && c>0 && a+b+c==180 ? "YES" : "NO"));}return 0;
}

C. Memory is Full

贪心，先移除占用内存较大的软件，复杂度O(nlogn)。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
int a[105];
int main()
{int T;scanf("%d",&T);while(T--){int k,m,n;scanf("%d%d%d",&k,&m,&n);int sum=0;for(int i=1;i<=n;i++){scanf("%d",&a[i]);sum+=a[i];}int emp=k-sum;if(emp>=m){printf("0\n");continue;}sort(a+1,a+n+1,greater<int>());for(int i=1;i<=n;i++){emp+=a[i];if(emp>=m){printf("%d\n",i);break;}}}return 0;
}

D. Sudoku

直接模拟，复杂度O(1)。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
char s[15][15];
bool vis[15];
int main()
{int T;scanf("%d",&T);while(T--){for(int i=0;i<9;i++)scanf("%s",s[i]);bool isok=1;for(int i=0;i<9;i++){bool flag=1;memset(vis,0,sizeof(vis));for(int j=0;j<9;j++)vis[s[i][j]-'1']=1;for(int j=0;j<9;j++)flag&=vis[j];isok&=flag;}for(int i=0;i<9;i++){bool flag=1;memset(vis,0,sizeof(vis));for(int j=0;j<9;j++)vis[s[j][i]-'1']=1;for(int j=0;j<9;j++)flag&=vis[j];isok&=flag;}for(int i=0;i<9;i+=3)for(int j=0;j<9;j+=3){memset(vis,0,sizeof(vis));for(int p=i;p<i+3;p++)for(int q=j;q<j+3;q++)vis[s[p][q]-'1']=1;bool flag=1;for(int p=0;p<9;p++)flag&=vis[p];isok&=flag;}printf("%s\n",(isok ? "Valid" : "Invalid"));}return 0;
}

E. Time Limit Exceeded?

记下每个数出现的次数，维护前缀和后可以O(1)计算每个数作为较小数时的贡献，复杂度O(n)。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
int cnt[10005],pre[10005];
int main()
{int T;scanf("%d",&T);while(T--){int n;scanf("%d",&n);int in;memset(cnt,0,sizeof(cnt));for(int i=1;i<=n;i++){scanf("%d",&in);cnt[in]++;}for(int i=1;i<=10000;i++)pre[i]=pre[i-1]+cnt[i];int ans=0;for(int i=1;i<=10000;i++){ans+=cnt[i]*(cnt[i]-1)/2;ans+=cnt[i]*(pre[min(10000,i+31)]-pre[i]);}printf("%d\n",ans);}return 0;
}

F. Palindrome

若s[i]与s[j]相同，则将s[i]与s[j]合并到同一集合中，并选取集合的代表元素，可以利用并查集维护，

在选取代表元素时，若集合中只有"?"，则选"?"为代表元素，否则任意选取一个不为"?"的字符，

完成合并过程后，逐个检查元素及其代表元素是否一致，并统计元素全为"?"的集合个数，复杂度O(nlogn)。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
const int MAXN=50005;
const int Mod=1000000007;
int fast_pow(int a,int k)
{int res=1;while(k>0){if(k&1)res=1LL*res*a%Mod;a=1LL*a*a%Mod;k>>=1;}return res;
}
char s[MAXN];
int p[MAXN];
bool vis[MAXN];
void Init(int n)
{for(int i=0;i<n;i++)p[i]=i;
}
int Find(int x)
{return x==p[x] ? x : p[x]=Find(p[x]);
}
void Union(int x,int y)
{x=Find(x);y=Find(y);if(x==y)return;if(s[x]=='?')p[x]=y;else p[y]=x;
}
int main()
{int T;scanf("%d",&T);while(T--){int n,m;scanf("%d%d",&n,&m);scanf("%s",s);Init(n);for(int i=0;i<n/2;i++)Union(i,n-1-i);int x,y;while(m--){scanf("%d%d",&x,&y);Union(x-1,y-1);}bool isok=1;int cnt=0;memset(vis,0,sizeof(vis));for(int i=0;i<n;i++){int t=Find(i);if(s[t]=='?'){if(!vis[t]){vis[t]=1;cnt++;}}else isok&=(s[t]==s[i] || s[i]=='?');}if(isok)printf("%d\n",fast_pow(26,cnt));else printf("0\n");}return 0;
}

G. Training Camp

状压dp，用dp[s]表示选取了s集合的最大收益，

根据拓扑序检验状态合法性并进行转移即可，复杂度O(n*2^n)。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<string>
#include<vector>
#include<map>
using namespace std;
map<string,int>mp;
string s[25];
int w[25],dp[1<<18];
vector<pair<int,int> >e;
bool used[25],in[25];
void init()
{e.clear();mp.clear();
}
int main()
{ios::sync_with_stdio(false);stringstream ss;int T;cin>>T;while(T--){int n,m;cin>>n>>m;string str;getline(cin,str);init();for(int i=1;i<=n;i++){w[i]=0;getline(cin,str);ss.clear();ss.str(str);string name="",word="";while(ss>>word){if(word[0]>='0' && word[0]<='9')for(int j=0;j<word.size();j++)w[i]=w[i]*10+word[j]-'0';else name+=word+" ";}mp[name]=i;}/*for(int i=1;i<=n;i++)cout<<s[i]<<" "<<w[i]<<endl;*/for(int i=1;i<=m;i++){getline(cin,str);ss.clear();ss.str(str);string name="",word="";int st,ed;while(ss>>word){if(word[0]=='-'){st=mp[name];name.clear();}else name+=word+" ";}ed=mp[name];e.push_back(make_pair(st,ed));}memset(dp,0,sizeof(dp));for(int mask=0;mask<(1<<n);mask++){memset(used,0,sizeof(used));int day=1;for(int i=0;i<n;i++)if(mask&(1<<i)){day++;used[i+1]=1;}bool isok=1;for(int i=0;i<e.size();i++)if(!used[e[i].first] && used[e[i].second]){isok=0;break;}if(!isok)continue;memset(in,0,sizeof(in));for(int i=0;i<e.size();i++)if(!used[e[i].first])in[e[i].second]=1;for(int i=0;i<n;i++)if(!used[i+1] && !in[i+1])dp[mask|(1<<i)]=max(dp[mask|(1<<i)],dp[mask]+day*w[i+1]);}cout<<dp[(1<<n)-1]<<endl;}return 0;
}

H. Capital City

双连通分量缩点，对得到的树进行两次bfs得到直径，然后枚举直径上的点求出重心，

考虑到可能有重边以及自环，需要进行去重，

去重时利用并查集维护，如果两个点之间有多条路径，则将这两个点合并为一个点，

由于要输出最小的标号，还需要选取每个集合中标号最小的点作为代表元素，复杂度O(nlogn+mlogm)。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<set>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
const int MAXN=100005;
const int MAXM=400005;
const int INF=0x3f3f3f3f;
//BCC
struct Edge
{int to,next,cost;bool cut;
}edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
int Get[MAXN];
int Index,top;
int block;
bool Instack[MAXN];
int bridge;
void addedge(int u,int v,int cost)
{edge[tot].to=v;edge[tot].next=head[u];edge[tot].cut=0;edge[tot].cost=cost;head[u]=tot++;
}
void Tarjan(int u,int pre)
{int v;Low[u]=DFN[u]=++Index;Stack[top++]=u;Instack[u]=1;for(int i=head[u];i!=-1;i=edge[i].next){v=edge[i].to;if(v==pre)continue;if(!DFN[v]){Tarjan(v,u);if(Low[u]>Low[v])Low[u]=Low[v];if(Low[v]>DFN[u]){bridge++;edge[i].cut=1;edge[i^1].cut=1;}}else if(Instack[v] && Low[u]>DFN[v])Low[u]=DFN[v];}if(Low[u]==DFN[u]){block++;do{v=Stack[--top];Instack[v]=0;Belong[v]=block;}while(v!=u);}
}
void init()
{tot=0;memset(head,-1,sizeof(head));
}
void get_BCC(int n)
{memset(DFN,0,sizeof(DFN));memset(Instack,0,sizeof(Instack));Index=top=block=bridge=0;Tarjan(1,0);memset(Get,INF,sizeof(Get));for(int i=1;i<=n;i++)Get[Belong[i]]=min(Get[Belong[i]],i);
}
//DSU
int p[MAXN];
void DSU_Init(int n)
{for(int i=1;i<=n;i++)p[i]=i;
}
int DSU_Find(int x)
{return x==p[x] ? x : p[x]=DSU_Find(p[x]);
}
void DSU_Union(int x,int y)
{x=DSU_Find(x);y=DSU_Find(y);if(x==y)return;p[x]=y;
}
void DSU_Compress(int n)
{for(int i=1;i<=n;i++)p[i]=DSU_Find(i);
}
//InputData
struct Input
{int u,v,cost;Input(){}Input(int _u,int _v,int _cost){if(_u>_v)swap(_u,_v);u=_u;v=_v;cost=_cost;}bool operator < (const Input &t)const{return v==t.v ? u<t.u : v<t.v;}
};
set<Input>s;
int vis[MAXN];
int Stand[MAXN];
//Tree
vector<pair<int,int> >e[MAXN];
void Prepare(int n)
{for(int i=1;i<=n;i++)e[i].clear();for(int i=1;i<=n;i++)for(int j=head[i];j!=-1;j=edge[j].next)if(edge[j].cut){e[Belong[i]].push_back(make_pair(Belong[edge[j].to],edge[j].cost));e[Belong[edge[j].to]].push_back(make_pair(Belong[i],edge[j].cost));}
}
pair<int,int> pre[MAXN];
ll dis[MAXN];
int BFS(int st)
{for(int i=1;i<=block;i++)dis[i]=(1LL<<62)-1;queue<int>q;q.push(st);pre[st].first=pre[st].second=-1;dis[st]=0;while(!q.empty()){int u=q.front();q.pop();for(int i=0;i<e[u].size();i++){int v=e[u][i].first;int c=e[u][i].second;if(dis[v]>dis[u]+c){q.push(v);dis[v]=dis[u]+c;pre[v]=make_pair(u,c);}}}int loc=st;ll Max=0;for(int i=1;i<=block;i++)if(dis[i]>Max){loc=i;Max=dis[i];}return loc;
}
//Main
int main()
{int T;scanf("%d",&T);while(T--){init();int n,m;scanf("%d%d",&n,&m);DSU_Init(n);s.clear();int a,b,c;for(int i=1;i<=m;i++){scanf("%d%d%d",&a,&b,&c);if(a==b)continue;Input t=Input(DSU_Find(a),DSU_Find(b),c);if(s.find(t)==s.end())s.insert(t);else DSU_Union(a,b);}DSU_Compress(n);int cnt=0;memset(Stand,INF,sizeof(Stand));memset(vis,0,sizeof(vis));for(int i=1;i<=n;i++){int t=DSU_Find(i);if(!vis[t])vis[t]=++cnt;Stand[vis[t]]=min(Stand[vis[t]],i);}set<Input>::iterator itr;for(itr=s.begin();itr!=s.end();itr++){Input t=*itr;int t1=DSU_Find(t.u);int t2=DSU_Find(t.v);if(t1==t2)continue;else{addedge(vis[t1],vis[t2],t.cost);addedge(vis[t2],vis[t1],t.cost);}}get_BCC(cnt);Prepare(cnt);int end1=BFS(1);int end2=BFS(end1);ll len=dis[end2];stack<pair<int,ll> >sst;ll now=0;sst.push(make_pair(end2,max(now,len-now)));for(int i=end2;i!=end1 && i!=-1;i=pre[i].first){now+=pre[i].second;ll t=max(now,len-now);while(!sst.empty() && t<sst.top().second)sst.pop();if(sst.empty() || t==sst.top().second)sst.push(make_pair(pre[i].first,t));}int ans_loc=INF;ll ans_pat=sst.top().second;while(!sst.empty()){ans_loc=min(ans_loc,Stand[Get[sst.top().first]]);sst.pop();}printf("%d %I64d\n",ans_loc,ans_pat);}return 0;
}

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