POJ 1149 PIGS（最大流）

POJ 1149 PIGS

题目链接

题意：有n个猪圈，m个顾客，猪圈中一开始有一些猪，顾客轮流来（注意是有先后顺序的），然后每个顾客会开启一些猪圈，在开启的猪圈中最多买b只猪，之后可以任意把剩下的猪分配到开着的猪圈中，问最多能卖出几只猪

思路：这题的关键在于建模，由于顾客有先后顺序，假如后来的顾客会开启x门，前面一个顾客也会开启x门，那么前面顾客相当与可以分配给后面顾客，
所以建模的方式为，源点和每个猪圈连，容量为猪圈猪数，每个猪圈和第一个开的顾客连，如果后面有顾客会开这个猪圈，则和之前的顾客连边，然后每个顾客在连向汇点，容量为会买的猪数

代码：

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int MAXNODE = 1205;const int MAXEDGE = 100005;typedef int Type;const Type INF = 0x3f3f3f3f;struct Edge {int u, v;Type cap, flow;Edge() {}Edge(int u, int v, Type cap, Type flow) {this->u = u;this->v = v;this->cap = cap;this->flow = flow;}};struct Dinic {int n, m, s, t;Edge edges[MAXEDGE];int first[MAXNODE];int next[MAXEDGE];bool vis[MAXNODE];Type d[MAXNODE];int cur[MAXNODE];vector<int> cut;void init(int n) {this->n = n;memset(first, -1, sizeof(first));m = 0;}void add_Edge(int u, int v, Type cap) {edges[m] = Edge(u, v, cap, 0);next[m] = first[u];first[u] = m++;edges[m] = Edge(v, u, 0, 0);next[m] = first[v];first[v] = m++;}bool bfs() {memset(vis, false, sizeof(vis));queue<int> Q;Q.push(s);d[s] = 0;vis[s] = true;while (!Q.empty()) {int u = Q.front(); Q.pop();for (int i = first[u]; i != -1; i = next[i]) {Edge& e = edges[i];if (!vis[e.v] && e.cap > e.flow) {vis[e.v] = true;d[e.v] = d[u] + 1;Q.push(e.v);}}}return vis[t];}Type dfs(int u, Type a) {if (u == t || a == 0) return a;Type flow = 0, f;for (int &i = cur[u]; i != -1; i = next[i]) {Edge& e = edges[i];if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {e.flow += f;edges[i^1].flow -= f;flow += f;a -= f;if (a == 0) break;}}return flow;}Type Maxflow(int s, int t) {this->s = s; this->t = t;Type flow = 0;while (bfs()) {for (int i = 0; i < n; i++)cur[i] = first[i];flow += dfs(s, INF);}return flow;}void MinCut() {cut.clear();for (int i = 0; i < m; i += 2) {if (vis[edges[i].u] && !vis[edges[i].v])cut.push_back(i);}}} gao;const int N = 1005;const int M = 105;int n, m, pre[N], vis[M][M];int main() {while (~scanf("%d%d", &n, &m)) {memset(pre, 0, sizeof(pre));gao.init(n + m + 2);int w;for (int i = 1; i <= n; i++) {scanf("%d", &w);gao.add_Edge(0, i, w);}int a, k, b;for (int i = 1; i <= m; i++) {scanf("%d", &a);while (a--) {scanf("%d", &k);if (!pre[k]) {pre[k] = i;gao.add_Edge(k, n + i, INF);} else {if (vis[pre[k]][i] == 0) {vis[pre[k]][i] = 1;gao.add_Edge(n + pre[k], n + i, INF);}}}scanf("%d", &b);gao.add_Edge(n + i, n + m + 1, b);}printf("%d\n", gao.Maxflow(0, n + m + 1));}return 0;}