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;}