[最大流] poj 2584：T-Shirt Gumbo

大致题意：
    已知n个同学和5种衣服，要让每一个人都有衣服穿。已知每个同学可以穿的衣服种类，每种衣服的数量。求衣服的数量能否满足同学的需求。

大致思路：
    标准的二分图多重匹配，设超级源汇点，超级源点向每个同学连边，容量都为1。每个同学都向他需要的衣服连边，容量也是1。每件衣服向汇点连边，容量为其数量。对这个图求出最大流，如果得到的值等于n则表示可以提供给所有人衣服，否则就是无法提供。
（在家里刷了3天终于刷到水题了，擦~~睡个好觉吧）

详细代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int inf=1<<30;
const int nMax=10005;
const int mMax=40000;

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
   // cout<<u<<" add "<<v<<" "<<w<<endl;
    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 dir[300];
int main(){
    char str[30];
    int n,i,j,s,t,m;
    dir['S']=1;dir['M']=2;dir['L']=3;dir['X']=4;dir['T']=5;//S M L X T
    while(scanf("%s",str)!=EOF&&strcmp(str,"ENDOFINPUT")!=0){
        scanf("%d",&n);
        ne=2;
        s=0,t=n+6;
        memset(head,0,sizeof(head));
        for(i=1;i<=n;i++){
            scanf("%s",str);
            addedge(s,i,1);
            for(j=dir[str[0]];j<=dir[str[1]];j++){
                addedge(i,j+n,1);
            }
        }
        for(i=1;i<=5;i++){
            scanf("%d",&m);
            addedge(i+n,t,m);
        }
        if(dinic(s,t)!=n)printf("I'd rather not wear a shirt anyway...\n");
        else printf("T-shirts rock!\n");
        scanf("%s", str);
    }
    return 0;
}