UVA 11045 My T-shirt suits me

又是一个需要拆分点的最大流题目，N件T恤，共种大小，每种数目一致为N/6,分给M个人，其中每个人可以选6个尺寸中的一种，问最后能否每个人分到一件衣服。

分析：

对每个人1~M，建立一条从源点0出发的边，容量为1，然后将每种尺寸拆分为两个点，对每个人分别建立到可选的两个尺寸的边，容量为inf，每个尺寸拆分后建立边的容量为衣服数目N/6,最后每个尺寸建立到汇点的无穷容量的边。

代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <iostream>
#include <cctype>
#include <queue>
#include <algorithm>
#include <cmath>
#define esp 1e-6
#define inf 0x0f0f0f0f
#define print(a) printf("%d\n", (a));
#define bug  puts("(****");
#define judge  freopen("in.txt","r",stdin);
using namespace std;

#define N 100
#define M 1000

char *si[]={"XXL","XL","L","M","S","XS"};
struct EDGE
{
    int i, c;
    EDGE *next, *ani;
} *Edge[N], E[M];
int Dfn[N], Now[N], cnt;
int src, sink;
int n, m;
int f(char *s)
{
    for(int i = 0; i < 6; i++)
        if(!strcmp(s,si[i]))
        return i;
}
void add(int i, int j, int c, EDGE &e1, EDGE &e2)
{
    e1.i = j, e1.c = c, e1.next = Edge[i], e1.ani = &e2, Edge[i] = &e1;
    e2.i = i, e2.c = 0, e2.next = Edge[j], e2.ani = &e1, Edge[j] = &e2;
}

void init(void)
{
    memset(Edge, 0, sizeof(Edge));
    cnt = 0;
}
int ISAP(int s, int end, int flow)
{
    if(s == end)
        return flow;
    int i, tab = n -1, now = 0, vary;
    for(EDGE *p = Edge[s]; p && flow - now; p = p->next)
        if(p->c)
        {
            if(Dfn[s] == Dfn[i = p->i] + 1)
                vary = ISAP(i, end, min(flow - now, p->c)),
                p->c -= vary, p->ani->c += vary, now +=vary;
            if(p->c)
                tab = min(tab, Dfn[i]);
            if(Dfn[src] == n)
                return now;
        }
    if(now == 0)
    {
        if(--Now[Dfn[s]] == 0)
            Dfn[src] = n;
        Now[Dfn[s] = tab + 1]++;
    }
    return now;
}
int max_flow(int s, int end)
{
    memset(Dfn, 0, sizeof(Dfn));
    memset(Now, 0, sizeof(Now));
    Now[0]  =  n;
    int ret = 0;
    for(; Dfn[s] < n;)
        ret += ISAP(s, end, inf);
    return ret;
}

char a[10],b[10];

int main(void)
{
    int T;
    for(int t = scanf("%d", &T); t <= T; t++)
    {
        init();
        scanf("%d%d", &n, &m);
        int c = n /6;
        n = 14 + m;
        for(int  u = m + 1; u <= m + 6; u ++)
        {
            int v = u + 6;
            add(u, v, c, E[cnt], E[cnt + 1]);
            cnt += 2;
            add(v, m + 13, inf, E[cnt], E[cnt + 1]);
            cnt += 2;
        }
        for(int i = 1; i <= m; i++)
        {
            scanf("%s%s", a, b);
            int u = f(a);
            int v = f(b);
            u += m + 1;
            v += m + 1;
            add(i, u, inf, E[cnt], E[cnt + 1]);
            cnt += 2;
            add(i, v, inf, E[cnt], E[cnt + 1]);
            cnt += 2;
            add(0, i, 1, E[cnt], E[cnt + 1]);
            cnt += 2;
        }
        src = 0, sink = m + 13;
        int ans = max_flow(src, sink);
        puts(ans != m ? "NO" : "YES");
    }
    return 0;
}