题意：
有M个猪圈，每个猪圈里初始时有若干头猪。一开始所有猪圈都是关闭的。依次来了N个顾客，每个顾客分别会打开指定的几个猪圈，从中买若干头猪。每个顾客分别都有他能够买的数量的上限。每个顾客走后，他打开的那些猪圈中的猪，都可以被任意地调换到其它开着的猪圈里，然后所有猪圈重新关上。问总共最多能卖出多少头猪。（1 <= N <= 100, 1 <= M <= 1000）
解析：
因为是依次 所以对于当前第i个顾客 所对应的猪圈 向前 i - 1 个顾客种有对应这个猪圈的顾客的其它猪圈 就好啦
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #include <bitset> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define rb(a) scanf("%lf", &a) #define rf(a) scanf("%f", &a) #define pd(a) printf("%d ", a) #define plld(a) printf("%lld ", a) #define pc(a) printf("%c ", a) #define ps(a) printf("%s ", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 1e5 + 10, INF = 0x7fffffff; int n, m, s, t; vector<int> g[maxn], f[maxn]; int head[maxn], cur[maxn], vis[maxn], d[maxn], cnt, nex[maxn << 1]; struct node { int u, v, c; }Node[maxn << 1]; void add_(int u, int v, int c) { Node[cnt].u = u; Node[cnt].v = v; Node[cnt].c = c; nex[cnt] = head[u]; head[u] = cnt++; } void add(int u, int v, int c) { add_(u, v, c); add_(v, u, 0); } bool bfs() { queue<int> Q; mem(d, 0); Q.push(s); d[s] = 1; while(!Q.empty()) { int u = Q.front(); Q.pop(); for(int i = head[u]; i != -1; i = nex[i]) { int v = Node[i].v; if(!d[v] && Node[i].c > 0) { d[v] = d[u] + 1; Q.push(v); if(v == t) return 1; } } } return d[t] != 0; } int dfs(int u, int cap) { int ret = 0; if(u == t || cap == 0) return cap; for(int &i = cur[u]; i != -1; i = nex[i]) { int v = Node[i].v; if(d[v] == d[u] + 1 && Node[i].c > 0) { int V = dfs(v, min(cap, Node[i].c)); Node[i].c -= V; Node[i ^ 1].c += V; ret += V; cap -= V; if(cap == 0) break; } } if(cap > 0) d[u] = -1; return ret; } int Dinic() { int ans = 0; while(bfs()) { memcpy(cur, head, sizeof head); ans += dfs(s, INF); } return ans; } int main() { while(scanf("%d %d", &m, &n) != EOF) { mem(head, -1); cnt = 0; for(int i = 0; i < maxn; i++) g[i].clear(), f[i].clear(); int w, x, v; s = 0, t = n + m + 10; rap(i, 1, m) { rd(w); add(i, t, w); } rap(i, 1, n) { rd(x); rap(j, 1, x) { rd(v); add(m + i, v, INF); for(int k = 0; k < g[v].size(); k++) { for(int p = 0; p < f[g[v][k]].size(); p++) add(m + i, f[g[v][k]][p], INF); } g[v].push_back(i); f[i].push_back(v); } rd(w); add(s, m + i, w); } pd(Dinic()); } return 0; }