[状态压缩+最大流]hdoj 3605：Escape

大致题意：
    在世界末日，有n个人要去m个星球。给出每个人能去的星球和每个星球能容纳的人数。判断是否存在可行的安排方案。n (1 <= n <= 100000), m (1 <= m <= 10)

大致思路：
    以为是水题上来就直接套二分图多重匹配来做，结果被TLE到各种吐血~~ 。后来看了题解才明白，这里要用状态压缩。因为所有的人可以按照可以去的星球划分为2^m类人。这样不按照人头来建图而用人的种类来建图，点的规模会大大缩小。终于领会到二进制压缩的妙处了。

详细代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int inf=1<<30;
const int nMax=20000;
const int mMax=100000;

class node{
    public:
    int c,u,v,next;
};node edge[mMax];
int ne, head[nMax];
int cur[nMax], ps[nMax], dep[nMax];

void addedge(int u, int v,int w){   ////dinic邻接表加边
    edge[ne].u = u;
    edge[ne].v = v;
    edge[ne].c = w;
    edge[ne].next = head[u];
    head[u] = ne ++;
    edge[ne].u = v;
    edge[ne].v = u;
    edge[ne].c = 0;
    edge[ne].next = head[v];
    head[v] = ne ++;
}

int dinic(int s, int t){                       //  dinic
    int tr, res = 0;
    int i, j, k, f, r, top;
    while(1){
        memset(dep, -1, sizeof(dep));
        for(f = dep[ps[0]=s] = 0, r = 1; f != r;)
            for(i = ps[f ++], j = head[i]; j; j = edge[j].next)
                if(edge[j].c && dep[k=edge[j].v] == -1){
                    dep[k] = dep[i] + 1;
                    ps[r ++] = k;
                    if(k == t){
                        f = r; break;
                    }
                }
        if(dep[t] == -1) break;
        memcpy(cur, head, sizeof(cur));
        i = s, top = 0;
        while(1){
            if(i == t){
                for(tr =inf, k = 0; k < top; k ++)
                    if(edge[ps[k]].c < tr)
                        tr = edge[ps[f=k]].c;
                for(k = 0; k < top; k ++){
                    edge[ps[k]].c -= tr;
                    edge[ps[k]^1].c += tr;
                }
                i = edge[ps[top=f]].u;
                res += tr;
            }
            for(j = cur[i]; cur[i]; j = cur[i] =edge[cur[i]].next){
                if(edge[j].c && dep[i]+1 == dep[edge[j].v]) break;
            }
            if(cur[i]){
                ps[top ++] = cur[i];
                i = edge[cur[i]].v;
            }
            else{
                if(top == 0) break;
                dep[i] = -1;
                i = edge[ps[-- top]].u;
            }
        }
    }
    return res;
}
int sum[200];
int main(){
    int n,m,i,j,s,t,a,b,N;
    while(scanf("%d%d",&n,&m)!=EOF){
        ne=2;
        memset(head,0,sizeof(head));
        N=1<<m;
        memset(sum,0,sizeof(sum));
        s=N+m+1,t=N+m+2;
        for(i=1;i<=n;i++){
            a=0;
            for(j=1;j<=m;j++){
                scanf("%d",&b);
                a=(a<<1)+b;
            }
            sum[a]++;
        }
      //  cout<<"fuck\n";
        for(i=0;i<N;i++){
            addedge(s,i,sum[i]);
        }
        for(i=0;i<N;i++){
            a=i;
            b=0;
            while(a){
                if(a&1){
                    addedge(i,b+N,sum[i]);
                }
                a>>=1;
                b++;
            }
        }
        for(i=N;i<N+m;i++){
            scanf("%d",&a);
            addedge(i,t,a);
        }
        if(dinic(s,t)==n){
            puts("YES\n");
        }
        else{
            puts("NO\n");
        }
    }
    return 0;
}