POJ 3189 Steady Cow Assignment(最大流）

POJ 3189 Steady Cow Assignment

题目链接

题意：一些牛，每个牛心目中都有一个牛棚排名，然后给定每个牛棚容量，要求分配这些牛给牛棚，使得所有牛对牛棚的排名差距尽量小

思路：这种题的标准解法都是二分一个差值，枚举下界确定上界，然后建图判断，这题就利用最大流进行判断，值得一提的是dinic的效率加了减枝还是是卡着时间过的，这题理论上用sap或者二分图多重匹配会更好

代码：

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int MAXNODE = 1025;const int MAXEDGE = 200005;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 = 25;int n, b, g[N][M], s[M];bool build(int x, int y) {gao.init(n + b + 2);for (int i = 1; i <= n; i++) {gao.add_Edge(0, i, 1);for (int j = x; j <= y; j++)gao.add_Edge(i, g[i][j] + n, 1);}for (int i = 1; i <= b; i++)gao.add_Edge(i + n, n + b + 1, s[i]);return gao.Maxflow(0, n + b + 1) == n;}int main() {while (~scanf("%d%d", &n, &b)) {for (int i = 1; i <= n; i++)for (int j = 1; j <= b; j++)scanf("%d", &g[i][j]);for (int i = 1; i <= b; i++)scanf("%d", &s[i]);int ans = 100;for (int i = 1; i <= b; i++) {for (int j = b; j >= i; j--) {if (ans <= j - i + 1) continue;if (build(i, j)) ans = min(ans, j - i + 1);}}printf("%d\n", ans);}return 0;}