题目地址：洛谷：【P2891】[USACO07OPEN]吃饭Dining – 洛谷、BZOJ：Problem 1711. — [Usaco2007 Open]Dining吃饭、POJ：3281 — Dining、OpenJudge百练：OpenJudge – 3479:Dining

题目描述

Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others.
Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although he might not be able to stuff everybody, he wants to give a complete meal of both food and drink to as many cows as possible.
Farmer John has cooked F (1 ≤ F ≤ 100) types of foods and prepared D (1 ≤ D ≤ 100) types of drinks. Each of his N (1 ≤ N ≤ 100) cows has decided whether she is willing to eat a particular food or drink a particular drink. Farmer John must assign a food type and a drink type to each cow to maximize the number of cows who get both.
Each dish or drink can only be consumed by one cow (i.e., once food type 2 is assigned to a cow, no other cow can be assigned food type 2).
有F种食物和D种饮料，每种食物或饮料只能供一头牛享用，且每头牛只享用一种食物和一种饮料。现在有n头牛，每头牛都有自己喜欢的食物种类列表和饮料种类列表，问最多能使几头牛同时享用到自己喜欢的食物和饮料。（1 <= f <= 100, 1 <= d <= 100, 1 <= n <= 100）

输入输出格式

输入格式：
Line 1: Three space-separated integers: N, F, and D
Lines 2..N+1: Each line i starts with a two integers Fi and Di, the number of dishes that cow i likes and the number of drinks that cow i likes. The next Fi integers denote the dishes that cow i will eat, and the Di integers following that denote the drinks that cow i will drink.

输出格式：
Line 1: A single integer that is the maximum number of cows that can be fed both food and drink that conform to their wishes

输入输出样例

输入样例#1：

4 3 3
2 2 1 2 3 1
2 2 2 3 1 2
2 2 1 3 1 2
2 1 1 3 3

输出样例#1：

3

说明

One way to satisfy three cows is:
Cow 1: no meal
Cow 2: Food #2, Drink #2
Cow 3: Food #1, Drink #1
Cow 4: Food #3, Drink #3
The pigeon-hole principle tells us we can do no better since there are only three kinds of food or drink. Other test data sets are more challenging, of course.

题解

“三分图匹配”？！
考虑把饮料的点放左边，食物的点放右边，建图为源→喜欢的饮料→牛→喜欢的食物→汇这样跑最大流。
然后你WA了。
这是因为你没有限制牛这个点内流过的流量，有可能给一只牛配了超过一套食物+饮料。因此我们要对牛拆点限流。

代码

// Code by KSkun, 2018/4
#include <cstdio>
#include <cstring>

#include <algorithm>
#include <queue>

typedef long long LL;

inline char fgc() {
    static char buf[100000], *p1 = buf, *p2 = buf;
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF 
        : *p1++;
}

inline LL readint() {
    register LL res = 0, neg = 1;
    register char c = fgc();
    while(c < '0' || c > '9') {
        if(c == '-') neg = -1;
        c = fgc();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 + c - '0';
        c = fgc();
    }
    return res * neg;
}

const int MAXN = 405, MAXM = 100005, INF = 1e9;

struct Edge {
    int to, cap, nxt;
} gra[MAXM << 1];
int head[MAXN], tot;

inline void addedge(int u, int v, int cap) {
    gra[tot] = Edge {v, cap, head[u]}; head[u] = tot++;
    gra[tot] = Edge {u, 0, head[v]}; head[v] = tot++;
}

int level[MAXN];
std::queue<int> que;

inline bool bfs(int s, int t) {
    memset(level, -1, sizeof(level));
    level[s] = 0; que.push(s);
    while(!que.empty()) {
        int u = que.front(); que.pop();
        for(int i = head[u]; ~i; i = gra[i].nxt) {
            int v = gra[i].to;
            if(gra[i].cap > 0 && level[v] == -1) {
                level[v] = level[u] + 1;
                que.push(v);
            }
        }
    }
    return level[t] != -1;
}

int cur[MAXN];
bool vis[MAXN];

inline int dfs(int u, int t, int left) {
    if(u == t || !left) 
        return left;
    int flow = 0; vis[u] = true;
    for(int &i = cur[u]; ~i; i = gra[i].nxt) {
        int v = gra[i].to;
        if(gra[i].cap > 0 && !vis[v] && level[v] == level[u] + 1) {
            int d = dfs(v, t, std::min(left, gra[i].cap));
            if(d > 0) {
                left -= d; flow += d;
                gra[i].cap -= d; gra[i ^ 1].cap += d;
                if(!left) {
                    level[u] = -1;
                    return flow;
                }
            }
        }
    }
    return flow;
}

inline int dinic(int s, int t) {
    int flow = 0;
    while(bfs(s, t)) {
        memset(vis, 0, sizeof(vis));
        memcpy(cur, head, sizeof(head));
        int f;
        while(f = dfs(s, t, INF)) {
            flow += f;
        }
    }
    return flow;
}

int n, f, d, fi, di, t, S, T;

// 1~n n+1~2n cow
// 2n+1~2n+f food
// 2n+f+1~2n+f+d drink

int main() {
    memset(head, -1, sizeof(head));
    n = readint(); f = readint(); d = readint();
    S = 2 * n + f + d + 1; T = S + 1;
    for(int i = 1; i <= n; i++) {
        addedge(i, i + n, 1);
    }
    for(int i = 1; i <= f; i++) {
        addedge(S, 2 * n + i, 1);
    }
    for(int i = 1; i <= d; i++) {
        addedge(2 * n + f + i, T, 1);
    }
    for(int i = 1; i <= n; i++) {
        fi = readint(); di = readint();
        while(fi--) {
            t = readint();
            addedge(2 * n + t, i, 1);
        }
        while(di--) {
            t = readint();
            addedge(i + n, 2 * n + f + t, 1);
        }
    }
    printf("%d", dinic(S, T));
    return 0;
}