(中等) CF 555E Case of Computer Network，双连通+树。
Andrewid the Android is a galaxy-known detective. Now he is preparing a defense against a possible attack by hackers on a major computer network.
In this network are n vertices, some pairs of vertices are connected by m undirected channels. It is planned to transfer q important messages via this network, the i-th of which must be sent from vertex si to vertex di via one or more channels, perhaps through some intermediate vertices.
To protect against attacks a special algorithm was developed. Unfortunately it can be applied only to the network containing directed channels. Therefore, as new channels can't be created, it was decided for each of the existing undirected channels to enable them to transmit data only in one of the two directions.
Your task is to determine whether it is possible so to choose the direction for each channel so that each of the q messages could be successfully transmitted.
人生中第一道div 1 E题，值得纪念一下。
题目就是给一个无向图，然后给每条边一个方向，问能不能满足Q个条件，每个条件就是能不能从ai到bi。
其实不难的题目，然而没有想到双连通卡了一段时间。
双连通的子图一定能够构造出使其满足任意两点都能互相到，所以先把原图缩点成树。
树的话方法是对于每个条件 a->b，c=lca（a，b），然后给a-c线段的端点复制+1 -1， c-b同样，但是是维护的另一个数组。
这样如果对于一个点维护的两个值都为正，说明不可能，否则可以。
但是注意图可能不连通，12组数据就是这样的，被坑了一次。
代码比较乱，如下：
// ━━━━━━神兽出没━━━━━━ // ┏┓ ┏┓ // ┏┛┻━━━━━━━┛┻┓ // ┃ ┃ // ┃ ━ ┃ // ████━████ ┃ // ┃ ┃ // ┃ ┻ ┃ // ┃ ┃ // ┗━┓ ┏━┛ // ┃ ┃ // ┃ ┃ // ┃ ┗━━━┓ // ┃ ┣┓ // ┃ ┏┛ // ┗┓┓┏━━━━━┳┓┏┛ // ┃┫┫ ┃┫┫ // ┗┻┛ ┗┻┛ // // ━━━━━━感觉萌萌哒━━━━━━ // Author : WhyWhy // Created Time : 2015年10月11日 星期日 22时40分22秒 // File Name : B.cpp #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; const int MaxN=200005; const int MaxM=MaxN<<1; ////////////////////////////////////////////////////// namespace TU { struct Edge { int u; int to,next; }; Edge E[MaxM]; int head[MaxN],Ecou; int LOW[MaxN],DFN[MaxN]; int couBridge; bool bridge[MaxM],cutp[MaxN]; int add_block[MaxN]; // The number of bolcks that is added by deleting a point.(not the number of blocks of this new graph!!!) int Index; struct newEdge { int u,v,num; newEdge(int _u=0,int _v=0,int _num=0):u(_u),v(_v),num(_num) {}; bool operator < (const newEdge &a) const { if(u==a.u) return v<a.v; return u<a.u; } bool operator == (const newEdge &a) const { return u==a.u && v==a.v; } }; bool chongE[MaxM]; // Parallel edge. newEdge remE[MaxM]; void Tarjan(int u,int pre) { int v; int couSon=0; LOW[u]=DFN[u]=++Index; for(int i=head[u];i!=-1;i=E[i].next) { v=E[i].to; if(v==pre) // !!! continue; if(!DFN[v]) { ++couSon; Tarjan(v,u); if(LOW[v]<LOW[u]) LOW[u]=LOW[v]; if(DFN[u]<LOW[v] && !chongE[i]) { bridge[i]=1; bridge[i^1]=1; // ++couBridge; } if(u!=pre && DFN[u]<=LOW[v]) { cutp[u]=1; ++add_block[u]; } } else if(DFN[v]<LOW[u]) LOW[u]=DFN[v]; } if(u==pre && couSon>1) { cutp[u]=1; add_block[u]=couSon-1; } } void getCUTP(int n) { sort(remE,remE+Ecou); for(int i=1;i<Ecou;++i) if(remE[i]==remE[i-1]) chongE[remE[i].num]=chongE[remE[i-1].num]=1; for(int i=1;i<=n;++i) if(!DFN[i]) Tarjan(i,i); } void addEdge(int u,int v) { E[Ecou].u=u; E[Ecou].to=v; E[Ecou].next=head[u]; bridge[Ecou]=0; chongE[Ecou]=0; remE[Ecou]=newEdge(u,v,Ecou); head[u]=Ecou++; } void init(int n) { Ecou=couBridge=Index=0; for(int i=1;i<=n;++i) { head[i]=-1; cutp[i]=DFN[i]=add_block[i]=0; } } }; ////////////////////////////////// int belong[MaxN]; int vis[MaxN]; int N; int TN; int Q; namespace TREE { const int LOG=20; struct Edge { int next,to; }; Edge E[MaxN*2]; int head[MaxN],Ecou; int dep[MaxN]; int par[MaxN][LOG]; void init() { memset(head,-1,sizeof(head)); Ecou=0; } void addEdge(int u,int v) { E[Ecou].next=head[u]; E[Ecou].to=v; head[u]=Ecou++; } int que[MaxN]; int first,last; void BFS(int root,int VI) { int t,v; first=last=0; dep[root]=1; par[root][0]=root; que[last++]=root; vis[root]=VI; while(last-first) { t=que[first++]; for(int i=1;i<LOG;++i) par[t][i]=par[par[t][i-1]][i-1]; for(int i=head[t];i!=-1;i=E[i].next) { v=E[i].to; if(v==par[t][0]) continue; dep[v]=dep[t]+1; par[v][0]=t; que[last++]=v; vis[v]=VI; } } } int query(int u,int v) { if(dep[u]<dep[v]) swap(u,v); for(int det=dep[u]-dep[v],i=0;det;det>>=1,++i) if(det&1) u=par[u][i]; if(u==v) return u; for(int i=LOG-1;i>=0;--i) if(par[u][i]!=par[v][i]) { u=par[u][i]; v=par[v][i]; } return par[u][0]; } // ------------- int rem1[MaxN],rem2[MaxN]; int sum1[MaxN],sum2[MaxN]; bool tvis[MaxN]; bool dfs111(int u,int pre) { tvis[u]=1; sum1[u]=rem1[u]; sum2[u]=rem2[u]; for(int i=head[u];i!=-1;i=E[i].next) if(E[i].to!=pre) { if(dfs111(E[i].to,u)==0) return 0; sum1[u]+=sum1[E[i].to]; sum2[u]+=sum2[E[i].to]; } if(sum1[u]>0 && sum2[u]>0 && pre!=-1) return 0; return 1; } bool judge() { int a,b; int c; int VI=0; for(int i=1;i<=N;++i) if(!vis[i]) BFS(i,++VI); while(Q--) { scanf("%d %d",&a,&b); a=belong[a]; b=belong[b]; if(vis[a]!=vis[b]) return 0; if(a!=b) { c=query(a,b); ++rem1[a]; --rem1[c]; ++rem2[b]; --rem2[c]; } } for(int i=1;i<=N;++i) if(tvis[i]==0) if(!dfs111(i,-1)) return 0; return 1; } }; /////////////////////////////////////// int que[MaxN]; int first,last; void bfs(int u,int be) { first=last=0; que[last++]=u; belong[u]=be; while(last-first) { u=que[first++]; for(int i=TU::head[u];i!=-1;i=TU::E[i].next) if(TU::bridge[i]==0 && belong[TU::E[i].to]==0) { belong[TU::E[i].to]=be; que[last++]=TU::E[i].to; } } } void getbelong() { N=0; for(int i=1;i<=TN;++i) if(!belong[i]) bfs(i,++N); } void addEdge() { for(int i=0;i<TU::Ecou;++i) if(TU::bridge[i]) TREE::addEdge(belong[TU::E[i].u],belong[TU::E[i].to]); } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int M; int a,b; scanf("%d %d %d",&TN,&M,&Q); TU::init(TN); while(M--) { scanf("%d %d",&a,&b); TU::addEdge(a,b); TU::addEdge(b,a); } TU::getCUTP(TN); getbelong(); TREE::init(); addEdge(); if(TREE::judge()) puts("Yes"); else puts("No"); return 0; }
