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二分答案+网络最大流
#include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<queue> #include<algorithm> using namespace std; int N,M; const int maxn = 2000 + 10; const int INF = 0x7FFFFFFF; struct Edge { int from, to, cap, flow; Edge(int u, int v, int c, int f) :from(u), to(v), cap(c), flow(f){} }; vector<Edge>edges; vector<int>G[maxn]; bool vis[maxn]; int d[maxn]; int cur[maxn]; int n, m, s, t; int U[maxn*500],V[maxn*500]; int tot; int T[maxn]; char ss[maxn]; void init() { for(int i = 0; i < maxn; i++) G[i].clear(); edges.clear(); tot=0; memset(T,0,sizeof T); } void AddEdge(int from, int to, int cap) { edges.push_back(Edge(from, to, cap, 0)); edges.push_back(Edge(to, from, 0, 0)); int w = edges.size(); G[from].push_back(w - 2); G[to].push_back(w - 1); } 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) { edges[G[x][i]].flow+=f; edges[G[x][i] ^ 1].flow-=f; flow+=f; a-=f; if(a==0) break; } } if(!flow) d[x] = -1; return flow; } int dinic(int s, int t) { int flow = 0; while (BFS()) { memset(cur, 0, sizeof(cur)); flow += DFS(s, INF); } return flow; } void Input() { int SS; for(int i=1;i<=N;i++) { gets(ss); for(int ii=0;ss[ii];ii++) if(ss[ii]==‘ ‘) {SS=ii;break;} int num=0; for(int ii=SS+1;;ii++) { if(ss[ii]>=‘0‘&&ss[ii]<=‘9‘) num=num*10+ss[ii]-‘0‘; if(ss[ii]==‘ ‘||ss[ii]==‘\0‘) { U[tot]=i;V[tot]=N+num+1;tot++; T[num]++; num=0; } if(ss[ii]==‘\0‘) break; } } } void Solve() { int Ma=-INF; for(int i=0;i<M;i++) if(T[i]>Ma) Ma=T[i]; int Min=0,Max=Ma; int Mid=(Min+Max)/2; s=0;t=N+M+1; while(1) { for(int i = 0; i < maxn; i++) G[i].clear(); edges.clear(); for(int i=1;i<=N;i++) AddEdge(s,i,1); for(int i=0;i<tot;i++) AddEdge(U[i],V[i],INF); for(int i=N+1;i<=N+M;i++) AddEdge(i,t,Mid); if(dinic(s,t)<N) { Min=Mid+1; Mid=(Min+Max)/2; } else { Max=Mid; Mid=(Min+Max)/2; } if(Min==Max) break; } printf("%d\n",Min); } int main() { while(~scanf("%d%d",&N,&M)) { if(!N&&!M) break; scanf("\n"); init(); Input(); Solve(); } return 0; }
POJ 2289 Jamie's Contact Groups
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原文地址:http://www.cnblogs.com/zufezzt/p/4823033.html