POJ 2112 Optimal Milking(二分+最大流)
POJ 2112 Optimal Milking
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; }