HDU 4560 我是歌手（最大流）

题意：

有n个歌手和m个流派，问你能够符合要求的最大匹配数是多少。类似于二分图匹配，所以用最大流可以做。既然是最大流，那么拆点是肯定的。我们把流派拆成两个点，两个点之间流量为k，前面的是不擅长的点，后面是擅长的点。如果歌手擅长某个流派，就连擅长的点，流量为1，否则连不擅长的点，流量为1。然后源点向每个歌手连边，擅长的点向汇点连边。这两组边上的流量是待定的，这关系到如何求最大匹配数。我们可以二分这两组边上的流量，也就是二分最大匹配数。如果最大流量等于匹配书*n，那么这个是可以达成的，继续二分就可以求出答案。用的Dinic，链式前向星存图，居然在HDU上的排名在12，挺快的。

代码

////  Created by  CQU_CST_WuErli//  Copyright (c) 2015 CQU_CST_WuErli. All rights reserved.//// #include<bits/stdc++.h>#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <string>#include <vector>#include <list>#include <map>#include <queue>#include <stack>#include <set>#include <algorithm>#include <sstream>#define CLR(x) memset(x,0,sizeof(x))#define OFF(x) memset(x,-1,sizeof(x))#define MEM(x,a) memset((x),(a),sizeof(x))#define For_UVa if (kase!=1) cout << endl#define BUG cout << "I am here" << endl#define lookln(x) cout << #x << "=" << x << endl#define SI(a) scanf("%d",&a)#define SII(a,b) scanf("%d%d",&a,&b)#define SIII(a,b,c) scanf("%d%d%d",&a,&b,&c)#define rep(flag,start,end) for(int flag=start;flag<=end;flag++)#define Rep(flag,start,end) for(int flag=start;flag>=end;flag--)#define Lson l,mid,rt<<1#define Rson mid+1,r,rt<<1|1#define Root 1,n,1#define BigInteger bigntemplate <typename T> T gcd(const T& a,const T& b) {return b==0?a:gcd(b,a%b);}const int MAX_L=2005;// For BigIntegerconst int INF_INT=0x3f3f3f3f;const long long INF_LL=0x7fffffff;const int MOD=1e9+7;const double eps=1e-9;const double pi=acos(-1);typedef long long  ll;using namespace std;const int N=5e6;int pnt[N],head[1000],nxt[N],cap[N];int cnt;int n,m,l,k;int link[88][88];void add_edge(int u,int v,int f){    pnt[cnt]=v;nxt[cnt]=head[u];head[u]=cnt;    cap[cnt++]=f;}int iter[1000],level[1000];bool bfs(int s,int t){    OFF(level);    queue<int> q;    q.push(s);    level[s]=0;    while (!q.empty()){        int x=q.front();q.pop();        for (int i=head[x];~i;i=nxt[i]){            int v=pnt[i];            if (level[v]==-1 && cap[i]){                level[v]=level[x]+1;                q.push(v);            }        }    }    return level[t]!=-1;}int dfs(int u,int t,int Flow){    if (u==t || Flow==0) return Flow;    int left=Flow;    for (int i=iter[u];~i;i=nxt[i]){        int v=pnt[i];        if (level[v]==level[u]+1 && cap[i]){            int d=dfs(v,t,min(cap[i],left));            iter[u]=i;            cap[i]-=d;            cap[i^1]+=d;            left-=d;            if (!left) return Flow;         }    }    level[u]=-1;    return Flow-left;}int Dinic(int s,int t){    int Max_flow=0;    while (bfs(s,t)){        for (int i=s;i<=t;i++) iter[i]=head[i];        Max_flow+=dfs(s,t,INF_INT);    }    return Max_flow;}int main(){#ifdef LOCAL    freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);//  freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);#endif    for (int T_T,kase=SI(T_T);kase<=T_T;kase++) {        SII(n,m);SII(l,k);        CLR(link);        int source=0,sink=n+2*m+1;        rep(i,1,l) {            int u,v;            SII(u,v);            link[u][v]=1;        }        OFF(head);cnt=0;        // source->singer        rep(i,1,n) {            add_edge(source,i,k);            add_edge(i,source,0);        }        int tt=cnt;        // singer->song        rep(i,1,n) rep(j,1,m) {            if (link[i][j]) {                add_edge(i,n+m+j,1);                add_edge(n+m+j,i,0);            }            else {                add_edge(i,n+j,1);                add_edge(n+j,i,0);            }        }        int kk=cnt;        rep(i,1,m) {            add_edge(n+i,n+m+i,k);            add_edge(n+m+i,n+i,0);        }        int ss=cnt;        //        rep(i,1,m) {            add_edge(n+m+i,sink,k);            add_edge(sink,n+m+i,0);        }        int l=0,r=m;        int ans=0;//      cout << l << ' ' << r << endl;        while (l<=r) {            int mid=(l+r)/2;//          lookln(mid);            for (int i=0;i<tt;i+=2) {                cap[i]=mid;                cap[i^1]=0;            }            for (int i=tt;i<kk;i+=2) {                cap[i]=1;                cap[i^1]=0;            }            for (int i=kk;i<ss;i+=2) {                cap[i]=k;                cap[i^1]=0;            }            for (int i=ss;i<cnt;i+=2) {                cap[i]=mid;                cap[i^1]=0;            }            int tmp=Dinic(source,sink);//          lookln(tmp);            if (tmp==mid*n) {                ans=mid;                l=mid+1;            }            else r=mid-1;        }        cout << "Case " << kase << ": " << ans << endl;    }    return 0;}