题目描述：

n个学生，5种衣服。

每一个学生可以要5种衣服的某一些，但是最终只能选一件。

告诉你每种衣服现存的个数。

问你能不能满足所有的学生。

解题报告：

标准水题。

超级源点，和所有学生连接，容量1.

学生和能要的衣服连接容量1.

衣服和一个超级汇点连接，容量是现存数量。

看最大流是否等于学生人数即可。

代码如下：dinic模板，代码略长。

#include<iostream>
using namespace std;
char str[20];
int n;
#define Max 0x1fffffff
#define size 30
struct edge{int from, to, val, next;}e[500];
int v[size], que[size], dis[size], cnt, cur[size];
void insert(int from, int to, int va)
{
    e[cnt].from = from, e[cnt].to = to; e[cnt].val = va;
    e[cnt].next = v[from];v[from] = cnt++;
    e[cnt].from = to, e[cnt].to = from; e[cnt].val = 0;
    e[cnt].next = v[to];v[to] = cnt++;
}
bool bfs(int n, int s, int t)
{
    int head, tail, id;
    head = tail = 0; que[tail++] = s;
    memset(dis, -1, sizeof(int) * n);dis[s] = 0;
    while(head < tail) // bfs,得到顶点i的距s的最短距离dis[i]
        for(id = v[que[head++]]; id != -1; id = e[id].next)
            if (e[id].val > 0 && dis[e[id].to] == -1)
            {
                dis[e[id].to] = dis[e[id].from] + 1;
                que[tail++] = e[id].to;
                if (e[id].to == t) return true;
            }
    return false;
}

int Dinic(int n, int s, int t)
{
    int maxflow = 0, tmp, i;
    while(bfs(n, s, t))
    {
        int u = s, tail = 0;
        for(i = 0; i < n; i++) cur[i] = v[i];
        while(cur[s] != -1)
            if (u != t && cur[u] != -1 && e[cur[u]].val > 0 && dis[u] != -1 && dis[u] + 1 == dis[e[cur[u]].to])
            {que[tail++] = cur[u]; u = e[cur[u]].to;}
            else if (u == t)
            {
                for(tmp = Max, i = tail - 1; i >= 0; i--) tmp = min(tmp, e[que[i]].val);
                for(maxflow += tmp, i = tail - 1; i >= 0; i--)
                {
                    e[que[i]].val -= tmp;
                    e[que[i] ^ 1].val += tmp;
                    if (e[que[i]].val == 0) tail = i;
                }
                u = e[que[tail]].from;
            }
            else
            {
                while(tail > 0 && u != s && cur[u] == -1) u = e[que[--tail]].from;
                cur[u] = e[cur[u]].next;
            }
    }
    return maxflow;
}
int hash(char c)
{
    if (c == 'S') return 1;
    else if (c == 'M') return 2;
    else if (c == 'L') return 3;
    else if (c == 'X') return 4;
    else return 5;
}
int main()
{
    while(scanf("%s", str) && strcmp(str, "ENDOFINPUT") != 0)
    {
        scanf("%d", &n);
        memset(v, -1, sizeof(v)); cnt = 0;
        int s = 0, t = n + 5 + 1;
        for(int i = 1; i <= n ;i++)
        {
            insert(s, i, 1);
            scanf("%s", str);
            int from = hash(str[0]), to = hash(str[1]);
            for(int j = from; j <= to; j++)
                insert(i, n + j, 1);
        }
        for(int i = 1; i <= 5; i++)
        {
            int tmp; scanf("%d", &tmp);
            insert(i + n, t, tmp);
        }
        if (Dinic(t + 1, s, t) != n) printf("I'd rather not wear a shirt anyway...\n");
        else printf("T-shirts rock!\n");
        scanf("%s", str);
    }
    return 0;
}