POJ 2112 Optimal Milking, 二分, floyd, 二分图

http://poj.org/problem?id=2112

 

Optimal Milking

TimeLimit: 2000MS           MemoryLimit: 30000K

Case TimeLimit: 1000MS

题目大意：

现在有K个机器，C头牛，每头牛要使用一个机器，每个机器最多被M头牛使用。

机器标号为1~K，牛标号为K+1~K+C，用矩阵给出这K+C个结点之间各条边的距离。

现在要使每头牛有一台机器用，问所有牛要走的最长一条边的最短距离是多少。

 

Input

* Line 1: A single line withthree space-separated integers: K, C, and M. 

* Lines 2.. ...: Each of these K+C lines of K+C space-separated integersdescribes the distances between pairs of various entities. The input forms asymmetric matrix. Line 2 tells the distances from milking machine 1 to each ofthe other entities; line 3 tells the distances from machine 2 to each of theother entities, and so on. Distances of entities directly connected by a pathare positive integers no larger than 200. Entities not directly connected by apath have a distance of 0. ...(后面的废话省略)

Output

A single line with a singleinteger that is the minimum possible total distance for the furthest walkingcow. 

Sample Input

2 3 2

0 3 2 1 1

3 0 3 2 0

2 3 0 1 0

1 2 1 0 2

1 0 0 2 0

Sample Output

2

Source

USACO2003 U S Open

 

 

 

解法：

问最大的最小值，显然要二分答案。

先用floyd求最短路；再二分答案，用二分结果来建立二分图；然后用裸的增广路来检验这个二分图能否满足匹配要求即可。

建二分图时，将每个机器分成M个结点，即可化成一般的二分图。

 

596K 297MS

 

<span style="font-size:18px;">/*floyd,二分答案，二分图匹配。简洁明了 */#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int inf=250,Max=0x3f3f3f3f;int dis[inf][inf];//二分图匹配：bool map[30*15+10][inf];//将每个机器拆成m个点存储。map[i][j]表示第i个机器到第j头牛之间的边 bool vis[30*15+10];int use[30*15+10];int k,c,m,n;void floyd(){for (int r=1;r<=n;r++)for (int i=1;i<=n;i++)for (int j=1;j<=n;j++)dis[i][j]=min(dis[i][j],dis[i][r]+dis[r][j]);}void buildmap(int maxd){//建二分图 memset(map,false,sizeof(map));for (int i=1;i<=k;i++)for (int j=k+1;j<=n;j++)if (dis[i][j]<=maxd)for (int r=0;r<m;r++)//把每个机器拆成m个 map[r*k+i][j-k]=true;}bool dfs(int i){for (int j=1;j<=k*m;j++){if (map[j][i] && !vis[j]){vis[j]=true;if (!use[j] || dfs(use[j])){use[j]=i;return true;}}}return false;}bool match(){//增广路求二分图匹配 memset(use,0,sizeof(use));for (int i=1;i<=c;i++){memset(vis,false,sizeof(vis));if (!dfs(i)) return false;//只要有一个点不能匹配就算建图失败 }return true;}int main(){cin>>k>>c>>m;n=k+c;for (int i=1;i<=n;i++) for (int j=1;j<=n;j++){scanf("%d",&dis[i][j]);if (!dis[i][j]) dis[i][j]=Max;}floyd();//二分答案 int l=0,r=200*n,ans;while(l<=r){int mid=(l+r)>>1;buildmap(mid);//建二分图：二分图中只保留长度小于等于mid的边 if (match()) r=mid-1,ans=mid;//能够匹配说明：最长边的最小值不够小 else l=mid+1;}cout<<ans;return 0;} </span>