uva 11045 My T-shirt suits me （二分图匹配 最大流）

uva 11045 My T-shirt suits me

题目大意：有n件衣服（一定是6的倍数，六种尺码n / 6套），m个试穿者，每个试穿者都有两种合适的尺码（尺码一共有六种：XS， S， M， L， XL， XXL）。问是否所有试穿者都能找到合适的衣服。

解题思路：设置一个超级源点，连向所有的试穿者，容量为1。把相同的衣服，当成不同的，比如XS型号的衣服有三件，我们则把它分为编号为1， 1 + 6， 1 + 12三件衣服。这样所有的衣服连向一个超级汇点，容量为一。然后把顾客和相应尺寸的衣服连起来（记得是同尺寸的所有衣服）。最后求最大流。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <map>
#include <iostream>
#include <queue>
using namespace std;
typedef long long ll;
const int N = 305;
const int INF = 0x3f3f3f3f;
const int OF = 50;
int n, m, L, s, t, rec[N];
map<string, int> mp2;
string size[7] = {"", "XS", "S", "M", "L", "XL", "XXL"};
struct Edge{
    int from, to, cap, flow; 
};

vector<Edge> edges;
vector<int> G[N];

void init() {
    memset(rec, 0, sizeof(rec));
    for (int i = 0; i < N; i++) G[i].clear();
    edges.clear();
}
void addEdge(int from, int to, int cap, int flow) {
    edges.push_back((Edge){from, to, cap, 0});
    edges.push_back((Edge){to, from, 0, 0});
    int m = edges.size();
    G[from].push_back(m - 2);
    G[to].push_back(m - 1);
} 
void input() {
    mp2[size[1]] = 1;
    mp2[size[2]] = 2;
    mp2[size[3]] = 3;
    mp2[size[4]] = 4;
    mp2[size[5]] = 5;
    mp2[size[6]] = 6;
    for (int i = 1; i <= m; i++) {
        addEdge(0, i, 1, 0);
    }
    string a, b;
    for (int i = 1; i <= m; i++) {
        cin >> a >> b;  
        for (int j = 0; j < L; j++) {
            int pa = mp2[a] + OF + j * 6;
            addEdge(i, pa, 01, 0);
            if (!rec[pa]) {
                addEdge(pa, t, 1, 0);   
                rec[pa] = 1;
            }
            int pb = mp2[b] + OF + j * 6;
            addEdge(i, pb, 1, 0);
            if (!rec[pb]) {
                addEdge(pb, t, 1, 0);   
                rec[pb] = 1;
            }
        }
    }
}
int vis[N], d[N];
int BFS() {
    memset(vis, 0, sizeof(vis));
    queue<int> Q;
    Q.push(s);
    d[s] = 0;
    vis[s] = 1;
    while (!Q.empty()) {
        int u = Q.front(); Q.pop(); 
        for (int i = 0; i < G[u].size(); i++) {
            Edge &e = edges[G[u][i]];   
            if (!vis[e.to] && e.cap > e.flow) {
                vis[e.to] = 1;  
                d[e.to] = d[u] + 1;
                Q.push(e.to);
            }
        }
    }
    return vis[t];
}

int cur[N];
int DFS(int u, int a) {
    if (u == t || a == 0) return a;
    int flow = 0, f; 
    for (int &i = cur[u]; i < G[u].size(); i++) {
        Edge &e = edges[G[u][i]];
        if (d[u] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
            e.flow += f;    
            edges[G[u][i]^1].flow -= f;
            flow += f;
            a -= f;
            if (a == 0) break;
        }
    }
    return flow;
}
int MF() { //dinic算法求最大流
    int ans = 0;
    while (BFS()) {
        memset(cur, 0, sizeof(cur));    
        ans += DFS(s, INF);
    }
    return ans;
}
int main() {
    int T;
    scanf("%d", &T);    
    while (T--) {
        scanf("%d %d", &n, &m);  
        s = 0, t = 200;
        L = n / 6;
        init();
        input();
        int ans;
        ans = MF();
        if (ans == m) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}