P2891 [USACO07OPEN]吃饭Dining 最大流

\(\color{#0066ff}{ 题目描述 }\)

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）

$\color{#0066ff}{ 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.

\(\color{#0066ff}{输出格式}\)

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

\(\color{#0066ff}{输入样例}\)

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

\(\color{#0066ff}{输出样例}\)

3

\(\color{#0066ff}{数据范围与提示}\)

1 <= f <= 100, 1 <= d <= 100, 1 <= n <= 100

\(\color{#0066ff}{ 题解 }\)

最大流

图从左至右依次为s，食物，牛，牛，饮料，t

s向所有食物连容量为1的边，所有饮料向t连容量为1的边

每头牛之间连容量为1的边

然后食物与左边的牛，右边的牛与饮料，按照输入连容量为1的边

这样保证了题目的两个条件

因为可能会有一头牛同时喜欢多种食物和饮料，而我们只能算一次，一头牛最多有1的流量！

#include<bits/stdc++.h>
#define LL long long
LL in() {
    char ch; LL x = 0, f = 1;
    while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
    for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
    return x * f;
}
const int maxn = 1e4;
struct node {
    int to, dis;
    node *nxt, *rev;
    node(int to = 0, int dis = 0, node *nxt = NULL, node *rev = NULL)
        : to(to), dis(dis), nxt(nxt), rev(rev) {}
    void *operator new(size_t) {
        static node *S = NULL, *T = NULL;
        return (S == T) && (T = (S = new node[1024]) + 1024), S++;
    }
}*head[maxn], *cur[maxn];
int dep[maxn];
int n, s, t, na, nb;
void add(int from, int to, int dis) {
    head[from] = new node(to, dis, head[from], NULL);
}
void link(int from, int to, int dis) {
    add(from, to, dis), add(to, from, 0);
    (head[from]->rev = head[to])->rev = head[from];
}
bool bfs() {
    for(int i = s; i <= t; i++) dep[i] = 0, cur[i] = head[i];
    std::queue<int> q;
    q.push(s);
    dep[s] = 1;
    while(!q.empty()) {
        int tp = q.front(); q.pop();
        for(node *i = head[tp]; i; i = i->nxt) 
            if(!dep[i->to] && i->dis)
                dep[i->to] = dep[tp] + 1, q.push(i->to);
    }
    return dep[t];
}
int dfs(int x, int change) {
    if(x == t || !change) return change;
    int flow = 0, ls;
    for(node *i = cur[x]; i; i = i->nxt) {
        cur[x] = i;
        if(dep[i->to] == dep[x] + 1 && (ls = dfs(i->to, std::min(change, i->dis)))) {
            change -= ls;
            flow += ls;
            i->dis -= ls;
            i->rev->dis += ls;
            if(!change) break;
        }
    }
    return flow;
}
int dinic() {
    int flow = 0;
    while(bfs()) flow += dfs(s, 0x7fffffff);
    return flow;
}
//na n n nb
int main() {
    n = in(), na = in(), nb = in();
    s = 0, t = n + n + na + nb + 1;
    for(int i = 1; i <= na; i++) link(s, i, 1);
    for(int i = 1; i <= nb; i++) link(n + na + n + i, t, 1);
    for(int i = 1; i <= n; i++) link(na + i, na + n + i, 1);
    for(int i = 1; i <= n; i++) {
        int ka = in(), kb = in();
        while(ka --> 0) link(in(), na + i, 1);
        while(kb --> 0) link(na + n + i, n + n + na + in(), 1);
    }
    printf("%d\n", dinic());
    return 0;
}

标签：dep,int,0066ff,USACO07OPEN,change,Dining,ls,P2891,color
来源： https://www.cnblogs.com/olinr/p/10336647.html