HDU-4292 Food(最大流，拆点)

题目链接：HDU-4292 Food

题意

有$N$个人、$F$种食物、$D$种饮料，每种食物和饮料有一定数量，每个人对每种食物和饮料有数量为1的需求或无需求，问这$F$种食物和$D$种饮料最多能满足多少个人的需求。

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思路

一个人拆成两个结点，分为左部点和右部点，左部点向右部点连容量为1的边，表示一个人对答案的贡献最多为1；

源点向代表食物的结点连边，容量为对应的食物的数量，食物向对这种食物有需求的人的左部点连边，容量为1，表示需求为1；

人的右部点向有需求的饮料连边，容量为1，表示对这种饮料需求为1，饮料向汇点连边，容量为对应的饮料的数量，然后跑最大流即可。

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代码实现

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using std::queue;
const int INF = 0x3f3f3f3f, N = 1000, M = 200000;
int head[N], d[N];
int s, t, tot, maxflow;
struct Edge
{
    int to, cap, nex;
} edge[M];
queue<int> q;
void add(int x, int y, int z) {
    edge[++tot].to = y, edge[tot].cap = z, edge[tot].nex = head[x], head[x] = tot;
    edge[++tot].to = x, edge[tot].cap = 0, edge[tot].nex = head[y], head[y] = tot;
}
bool bfs() {
    memset(d, 0, sizeof(d));
    while (q.size()) q.pop();
    q.push(s); d[s] = 1;
    while (q.size()) {
        int x = q.front(); q.pop();
        for (int i = head[x]; i; i = edge[i].nex) {
            int v = edge[i].to;
            if (edge[i].cap && !d[v]) {
                q.push(v);
                d[v] = d[x] + 1;
                if (v == t) return true;
            }
        }
    }
    return false;
}
int dinic(int x, int flow) {
    if (x == t) return flow;
    int rest = flow, k;
    for (int i = head[x]; i && rest; i = edge[i].nex) {
        int v = edge[i].to;
        if (edge[i].cap && d[v] == d[x] + 1) {
            k = dinic(v, std::min(rest, edge[i].cap));
            if (!k) d[v] = 0;
            edge[i].cap -= k;
            edge[i^1].cap += k;
            rest -= k;
        }
    }
    return flow - rest;
}
void init(int v_num) {
    tot = 1, maxflow = 0;
    s = v_num, t = s + 1;
    memset(head, 0, sizeof(head));
}

int main() {
    int nn, nf, nd;
    while (~scanf("%d %d %d", &nn, &nf, &nd)) {
        init(nn * 2 + nf + nd);
        for (int i = 0, num; i < nf; i++) {
            scanf("%d", &num);
            add(s, i, num);
        }
        for (int i = 0, num; i < nd; i++) {
            scanf("%d", &num);
            add(i + nf, t, num);
        }
        char str[210];
        for (int i = 0; i < nn; i++) {
            scanf(" %s", str);
            for (int j = 0; j < nf; j++) {
                if (str[j] == 'Y') add(j, i + nf + nd, 1);
            }
            add(nf + nd + i, nf + nd + nn + i, 1);
        }
        for (int i = 0; i < nn; i++) {
            scanf(" %s", str);
            for (int j = 0; j < nd; j++) {
                if (str[j] == 'Y') add(nn + nf + nd + i, nf + j, 1);
            }
        }
        while (bfs()) maxflow += dinic(s, INF);
        printf("%d\n", maxflow);
    }
    return 0;
}

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