UVA 1660 Cable TV Network（最大流）

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题意：给一个n个点的无向图，求它的点连通度，即最少删除多少个点，使得图不连通

思路：求点连通度，边连通度都可以转化为网络流来做，拆点如果有U->V，那么连U+N->V和V+N->U,然后枚举源点汇点求最大流即可

#include<bits/stdc++.h>using namespace std;#define INF 1e9const int maxn = 1000;struct Edge{int from,to,cap,flow;Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}};struct Dinic{int n,m,s,t;vector<Edge>edges;vector<int> G[maxn];bool vis[maxn];int d[maxn];int cur[maxn];void AddEdge(int from,int to,int cap){edges.push_back(Edge(from,to,cap,0));edges.push_back(Edge(to,from,0,0));m = edges.size();G[from].push_back(m-2);G[to].push_back(m-1);}void init(int n){for (int i = 0;i<=n;i++)G[i].clear();edges.clear();}bool BFS(){memset(vis,0,sizeof(vis));queue<int>q;q.push(s);d[s]=0;vis[s]=1;while (!q.empty()){int x = q.front();q.pop();for (int i = 0;i<G[x].size();i++){Edge &e = edges[G[x][i]];if (!vis[e.to] && e.cap>e.flow){vis[e.to]=1;d[e.to]=d[x]+1;q.push(e.to);}}}return vis[t];}int DFS(int x,int a){if (x==t || a==0)return a;int flow = 0,f;for (int &i=cur[x];i<G[x].size();i++){Edge&e = edges[G[x][i]];if (d[x]+1==d[e.to] && (f=DFS(e.to,min(a,e.cap-e.flow)))>0){e.flow+=f;edges[G[x][i]^1].flow-=f;flow+=f;a-=f;if (a==0)break;}}return flow;}int Maxflow(int s,int t){this->s=s;this->t=t;int flow = 0;while (BFS()){memset(cur,0,sizeof(cur));flow+=DFS(s,INF);}return flow;}}di;int uu[maxn],vv[maxn];void addedge(int n,int m){di.init(2*n+2);for (int i = 1;i<=n;i++)di.AddEdge(i,i+n,1);for (int i = 0;i<m;i++){di.AddEdge(uu[i]+n,vv[i],INF);di.AddEdge(vv[i]+n,uu[i],INF);}}int main(){int n,m;while (scanf("%d%d",&n,&m)!=EOF){di.init(2*n+2);for (int i = 0;i<m;i++){            int u,v;scanf(" (%d,%d)",&uu[i],&vv[i]);uu[i]++,vv[i]++;}int ans = INF;for (int i = 1;i<=n;i++)for (int j = 1;j<=n;j++)if (i==j)continue;        else{                    addedge(n,m);ans = min(ans,di.Maxflow(i+n,j));}if (ans == INF)ans = n;printf("%d\n",ans);}}

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