题意：n个男孩n个女孩，女孩选男孩，每个女孩都要选到不同的人k对女孩有相同选择标准，女孩每轮都选择没选过的男孩，问总共能选几轮。思路：女孩1..n，男孩n+1..2*n编号由女孩到男孩建容量为1的边起点st=2*n+1，到1..n建边；n+1..2*n到终点ed=2*n+2建边二分搜索最大容量即为答案。详见代码：/********************************************************* file name: hdu3081.cpp author : kereo create time: 2015年02月17日 星期二 22时23分44秒 *********************************************************/ #include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<set> #include<map> #include<vector> #include<stack> #include<cmath> #include<string> #include<algorithm> using namespace std; typedef long long ll; const int sigma_size=26; const int N=200+50; const int MAXN=100000+50; const int inf=0x3fffffff; const double eps=1e-8; const int mod=1000000000+7; #define L(x) (x<<1) #define R(x) (x<<1|1) #define PII pair<int, int> #define mk(x,y) make_pair((x),(y)) int n,m,k,edge_cnt,top; int fa[N]; int head[N],que[N],s[N],cur[N],gap[N],dep[N]; struct Edge{ int v,cap,flow,next; }edge[MAXN]; PII mp[MAXN]; void init(){ edge_cnt=top=0; memset(head,-1,sizeof(head)); } void addedge(int u,int v,int cap){ edge[edge_cnt].v=v; edge[edge_cnt].cap=cap; edge[edge_cnt].flow=0; edge[edge_cnt].next=head[u]; head[u]=edge_cnt++; edge[edge_cnt].v=u; edge[edge_cnt].cap=0; edge[edge_cnt].flow=0; edge[edge_cnt].next=head[v]; head[v]=edge_cnt++; } int findset(int x){ return fa[x] == x ? x : fa[x]=findset(fa[x]); } void Union(int x,int y){ int fx=findset(x); int fy=findset(y); if(fx == fy) return ; fa[fx]=fy; } void bfs(int st,int ed){ int front=0,rear=0; memset(gap,0,sizeof(gap)); memset(dep,-1,sizeof(dep)); dep[ed]=0; gap[0]=1; que[rear++]=ed; while(front!=rear){ int u=que[front++]; for(int i=head[u];i!=-1;i=edge[i].next){ int v=edge[i].v; if(dep[v]!=-1) continue; dep[v]=dep[u]+1; gap[dep[v]]++; que[rear++]=v; } } } int isap(int st,int ed){ bfs(st,ed); memcpy(cur,head,sizeof(head)); int ans=0,u=st; while(dep[st]<2*n+2){ if(u == ed){ int Min=inf,inser; for(int i=0;i<top;i++){ if(edge[s[i]].cap-edge[s[i]].flow<Min){ Min=edge[s[i]].cap-edge[s[i]].flow; inser=i; } } for(int i=0;i<top;i++) edge[s[i]].flow+=Min,edge[s[i]^1].flow-=Min; ans+=Min; top=inser; u=edge[s[top]^1].v; continue; } int flag=0,v; for(int i=cur[u];i!=-1;i=edge[i].next){ v=edge[i].v; if(edge[i].cap>edge[i].flow && dep[v]+1 == dep[u]){ flag=1; cur[u]=i; break; } } if(flag){ s[top++]=cur[u]; u=v; continue; } int d=2*n+2; for(int i=head[u];i!=-1;i=edge[i].next){ int v=edge[i].v; if(edge[i].cap>edge[i].flow && dep[v]<d){ cur[u]=i,d=dep[v]; } } gap[dep[u]]--; if(!gap[dep[u]]) return ans; dep[u]=d+1; gap[dep[u]]++; if(u!=st) u=edge[s[--top]^1].v; } return ans; } void Graph(int cap){ init(); int Map[N][N]; memset(Map,0,sizeof(Map)); for(int i=1;i<=n;i++){ addedge(0,i,cap); addedge(i+n,2*n+1,cap); } for(int i=1;i<=m;i++){ int u=mp[i].first; int v=mp[i].second; for(int j=1;j<=n;j++){ if(findset(u) == findset(j)){ if(Map[j][v]) continue; Map[j][v]=1; addedge(j,n+v,1); } } } } int main(){ int T; scanf("%d",&T); while(T--){ init(); scanf("%d%d%d",&n,&m,&k); for(int i=1;i<=n;i++) fa[i]=i; for(int i=1;i<=m;i++){ int u,v; scanf("%d%d",&u,&v); mp[i]=mk(u,v); } for(int i=0;i<k;i++){ int u,v; scanf("%d%d",&u,&v); Union(u,v); } int ans=0,l=0,r=n; while(l<=r){ int mid=(l+r)>>1; Graph(mid); int res=isap(0,2*n+1); if(res>=n*mid){ ans=mid; l=mid+1; } else r=mid-1; } printf("%d\n",ans); } return 0; }
hdu3081 Marriage Match II 二分+最大流
原文地址：http://blog.csdn.net/u011645923/article/details/43870207
