poj2122 Optimal Milking

Posted 2020-08-24 MashiroSky

http://poj.org/problem?id=2112 (题目链接)

题意

　　有K个能挤M头奶牛的挤奶机和C头奶牛，告诉一些挤奶机和奶牛间距离，求最优分配方案使最大距离最小。

Solution

　　先Floyd跑出两两点之间的最短距离，二分答案，最大流。

细节

　　注意距离不超过200是Floyd之前两点之间的距离不超过200。

代码

// poj2112
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define LL long long
#define inf 10000000
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;

const int maxn=500010;
struct edge {int to,next,w;}e[maxn<<1];
int head[maxn],d[maxn],f[300][300];
int n,m,cnt=1,es,et,C,K,M,ans;

void link(int u,int v,int w) {
	e[++cnt]=(edge){v,head[u],w};head[u]=cnt;
	e[++cnt]=(edge){u,head[v],w};head[v]=cnt;
}
void Floyd() {
	for (int k=1;k<=K+C;k++)
		for (int i=1;i<=K+C;i++)
			for (int j=1;j<=K+C;j++)
				f[i][j]=min(f[i][j],f[i][k]+f[k][j]);
}
bool bfs() {
	memset(d,-1,sizeof(d));
	queue<int> q;q.push(es);d[es]=0;
	while (!q.empty()) {
		int x=q.front();q.pop();
		for (int i=head[x];i;i=e[i].next) if (e[i].w && d[e[i].to]<0) {
				d[e[i].to]=d[x]+1;
				q.push(e[i].to);
			}
	}
	return d[et]>0;
}
int dfs(int x,int f) {
	if (x==et || f==0) return f;
	int w,used=0;
	for (int i=head[x];i;i=e[i].next) if (e[i].w && d[e[i].to]==d[x]+1) {
			w=dfs(e[i].to,min(e[i].w,f-used));
			used+=w;
			e[i].w-=w;e[i^1].w+=w;
			if (used==f) return used;
		}
	if (!used) d[x]=-1;
	return used;
}
bool Dinic(int mid) {
	int ans=0;memset(head,0,sizeof(head));
	cnt=1;
	for (int i=K+1;i<=K+C;i++) {
		for (int j=1;j<=K;j++) if (f[i][j]<=mid) link(i,j,1);
		link(es,i,1);
	}
	for (int i=1;i<=K;i++) link(i,et,M);
	while (bfs()) ans+=dfs(es,inf);
	return ans==C;
}
int main() {
	scanf("%d%d%d",&K,&C,&M);
	es=K+C+1;et=es+1;
	for (int i=1;i<=K+C;i++)
		for (int j=1;j<=K+C;j++) {
			scanf("%d",&f[i][j]);
			if (f[i][j]==0) f[i][j]=inf;
		}
	Floyd();
	int L=0,R=60000,res;
	while (L<=R) {
		int mid=(L+R)>>1;
		if (Dinic(mid)) res=mid,R=mid-1;
		else L=mid+1;
	}
	printf("%d",res);
	return 0;
}