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POJ 2112 Optimal Milking 二分答案+最大流
首先二分最长的边,然后删去所有比当前枚举的值长的边,算最大流,看是否能满足所有的牛都能找到挤奶的地方
#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <climits>#include <string>#include <iostream>#include <map>#include <cstdlib>#include <list>#include <set>#include <queue>#include <stack>using namespace std;typedef long long LL;const int maxn = 300;const int INF = INT_MAX / 4;int K,C,M,N,s,t;int dist[maxn][maxn];int cap[maxn][maxn],sig[maxn],vis[maxn][maxn];void floyd() { for(int k = 1;k <= N;k++) { for(int i = 1;i <= N;i++) { for(int j = 1;j <= N;j++) { dist[i][j] = min(dist[i][j],dist[i][k] + dist[k][j]); } } }}void build_graph(int minval) { memset(cap,0,sizeof(cap)); s = 0; t = N + 1; for(int i = K + 1;i <= N;i++) cap[s][i] = 1; for(int i = 1;i <= K;i++) cap[i][t] = M; for(int i = K + 1;i <= N;i++) { for(int j = 1;j <= K;j++) if(dist[i][j] <= minval) { cap[i][j] = 1; } }}int q[maxn * 2],qs,qe;bool bfs() { qs = 0; qe = 1; q[qs] = s; memset(sig,0,sizeof(sig)); sig[s] = 1; while(qs < qe) { int v = q[qs++]; if(v == t) break; for(int i = s;i <= t;i++) if(cap[v][i] && !sig[i]) { sig[i] = sig[v] + 1; q[qe++] = i; } } if(sig[t] == 0) return false; else return true;}int dinic(int now,int alpha) { int rest = alpha; if(now == t) return alpha; for(int i = s;i <= t;i++) {
//这里的限制条件一开始没加,效率非常低 if(sig[i] == sig[now] + 1 && rest && cap[now][i]) { int ret = dinic(i,min(rest,cap[now][i])); cap[now][i] -= ret; cap[i][now] += ret; rest -= ret; } } return alpha - rest;}bool ok(int val) { build_graph(val); int ans = 0; while(bfs()) { int ret = dinic(s,INF); ans += ret; } return ans >= C;}void solve() { floyd(); int mine = INF,maxe = -1; for(int i = 1;i <= N;i++) { for(int j = 1;j <= N;j++) { if(dist[i][j] != INF) { mine = min(mine,dist[i][j]); maxe = max(maxe,dist[i][j]); } } } int l = mine,r = maxe,ans; while(l <= r) { int mid = (l + r) / 2; if(ok(mid)) ans = mid,r = mid - 1; else l = mid + 1; } printf("%d\n",ans);}int main() { while(scanf("%d%d%d",&K,&C,&M) != EOF) { N = K + C; for(int i = 1;i <= N;i++) { for(int j = 1;j <= N;j++) { scanf("%d",&dist[i][j]); if(dist[i][j] == 0) dist[i][j] = INF; } } solve(); } return 0;}
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