POJ 2112 Optimal Milking
题目链接
题意：给定一些机器和奶牛，在给定距离矩阵，（不在对角线上为0的值代表不可达），每一个机器能容纳m个奶牛。问全部奶牛都能挤上奶，那么走的距离最大的奶牛的最小值是多少
思路：明显的二分+最大流。注意floyd求出的距离矩阵最大值可能不止200，所以二分的上限要注意
代码：
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;const int MAXNODE = 255;
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 = 255;int n, k, c, m, g[N][N];bool judge(int len) {gao.init(n + 2);for (int i = k + 1; i <= n; i++) gao.add_Edge(0, i, 1);for (int i = 1; i <= k; i++) {gao.add_Edge(i, n + 1, m);for (int j = k + 1; j <= n; j++) {if (g[j][i] > len) continue;gao.add_Edge(j, i, 1);}}return gao.Maxflow(0, n + 1) == c;
}int main() {while (~scanf("%d%d%d", &k, &c, &m)) {n = k + c;for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++) {scanf("%d", &g[i][j]);if (i != j && g[i][j] == 0)g[i][j] = INF;}int l = 0, r = 0;for (int k = 1; k <= n; k++) {for (int i = 1; i <= n; i++) {for (int j = 1; j <= n; j++) {g[i][j] = min(g[i][j], g[i][k] + g[k][j]);if (g[i][j] != INF) r = max(r, g[i][j]);}}}while (l < r) {int mid = (l + r) / 2;if (judge(mid)) r = mid;else l = mid + 1;}printf("%d\n", l);}return 0;
}
