题目来源:http://poj.org/problem?id=3281
网络流建图,最大流问题。
边的方向为:s->食物->牛->牛->饮料->t。
将牛拆成两个顶点,之间连一条容量为1的边,这样避免了一头牛杯分配多组食物和饮料的方案,计算出图中的最大流即可。
代码:
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
struct edge{
int to,next,cap,rev;
}e[100100];
bool vis[1010];
int head[3010],n,f,d,cnt=0;
void ins(int s,int t,int c) {
e[++cnt].to = t;
e[cnt].next = head[s];
head[s] = cnt;
e[cnt].cap = c;
e[cnt].rev = cnt + 1;
e[++cnt].to = s;
e[cnt].next = head[t];
head[t] = cnt;
e[cnt].cap = 0;
e[cnt].rev = cnt - 1;
}
int dfs(int v,int t,int f) {
if (v == t)return f;
vis[v] = 1;
for (int i = head[v]; i; i = e[i].next) {
if (!vis[e[i].to] && e[i].cap > 0) {
int d = dfs(e[i].to, t, min(f, e[i].cap));
if (d > 0) {
e[i].cap -= d;
e[e[i].rev].cap += d;
return d;
}
}
}
return 0;
}
int max_flow(int s,int t) {
int flow = 0;
for (;;) {
memset(vis, 0, sizeof(vis));
int f = dfs(s, t, 1e9);
if (f == 0)return flow;
flow += f;
}
}
int main() {
while (~scanf("%d%d%d", &n, &f, &d)) {
memset(e, 0, sizeof(e));
memset(head, 0, sizeof(head));
cnt = 0;
int s = f + d + 1 + 2 * n, t = f + d + 2 + 2 * n;
int x, y;
for (int i = 1; i <= f; ++i)
ins(s, i, 1);
for (int i = 1; i <= d; ++i)
ins(f + 2 * n + i, t, 1);
for (int i = 1; i <= n; ++i)
ins(f + i, f + i + n, 1);
for (int i = 1; i <= n; ++i) {
scanf("%d%d", &x, &y);
int num;
for (int j = 1; j <= x; ++j) {
scanf("%d", &num);
ins(num, i + f, 1);
}
for (int j = 1; j <= y; ++j) {
scanf("%d", &num);
ins(f + n + i, f + n * 2 + num, 1);
}
}
printf("%d\n", max_flow(s, t));
}
return 0;
}