2112 Optimal Milking //MaxMatch

Optimal Milking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 3639		Accepted: 1458
Case Time Limit: 1000MS

Description

FJ has moved his K (1 <= K <= 30) milking machines out into the cow pastures among the C (1 <= C <= 200) cows. A set of paths of various lengths runs among the cows and the milking machines. The milking machine locations are named by ID numbers 1..K; the cow locations are named by ID numbers K+1..K+C.  

Each milking point can "process" at most M (1 <= M <= 15) cows each day.  

Write a program to find an assignment for each cow to some milking machine so that the distance the furthest-walking cow travels is minimized (and, of course, the milking machines are not overutilized). At least one legal assignment is possible for all input data sets. Cows can traverse several paths on the way to their milking machine.  

Input

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

* Lines 2.. ...: Each of these K+C lines of K+C space-separated integers describes the distances between pairs of various entities. The input forms a symmetric matrix. Line 2 tells the distances from milking machine 1 to each of the other entities; line 3 tells the distances from machine 2 to each of the other entities, and so on. Distances of entities directly connected by a path are positive integers no larger than 200. Entities not directly connected by a path have a distance of 0. The distance from an entity to itself (i.e., all numbers on the diagonal) is also given as 0. To keep the input lines of reasonable length, when K+C > 15, a row is broken into successive lines of 15 numbers and a potentially shorter line to finish up a row. Each new row begins on its own line.  

Output

A single line with a single integer that is the minimum possible total distance for the furthest walking cow.  

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

USACO 2003 U S Open

 

 

 

 

二分图的多重匹配问题,可以先用Floyd算法计算出每两点之间的最短距离,然后找出这些距离中的最大值也就是那条最长路径high,在区间[0,high]中用二分法不断二分得到mid,根据mid来建立二分图,因为这里假设mid是最长路径的最小值,所以当用Floyd算大求解出来的d[i][j]<=mid时,就可以建立一条边,然后根据建立起来的二分图,判断其是否能满足多重匹配.

 

#include<stdio.h>
#include<string.h>
const int INF=1<<25-1;
int mat[250][250];
bool map[205][35];
int link[35][20];
bool usedif[35];
int k,c,m,high;
void floyed()
{
    for(int t=1; t<=k+c; t++)
        for(int i=1; i<=k+c; i++)
            for(int j=1; j<=k+c; j++)
                if(mat[i][t]+mat[t][j]<mat[i][j]) mat[i][j]=mat[i][t]+mat[t][j];
}
bool can(int t)
{
    for(int i=1; i<=k; i++)
    {
        if(usedif[i]==0&&map[t][i])
        {
            usedif[i]=1;
            if(link[i][0]<m)
            {
                link[i][++link[i][0]]=t;
                return true;
            }
            else
            {
                for(int j=1; j<=link[i][0]; j++)
                    if(can(link[i][j]))
                    {
                        link[i][j]=t;
                        return true;
                    }
            }
        }
    }
    return false;
}

bool check(int limit)
{
    memset(map,false,sizeof(map));
    for(int i=1; i<=k; i++)  link[i][0]=0;
    for(int i=1; i<=k; i++)
        for(int j=k+1; j<=k+c; j++)
            if(mat[i][j]<=limit)  map[j-k][i]=true;
    for(int i=1; i<=c; i++)
    {
        memset(usedif,0,sizeof(usedif));
        if(!can(i))  return false;
    }
    return true;
}

int find()
{
    int high=-1,low=0;
    for(int i=1; i<=k+c; i++)
        for(int j=1; j<=k+c; j++)
            if(mat[i][j]!=INF&&mat[i][j]>high) high=mat[i][j];
    while(low<high)
    {
        int mid=(low+high)>>1;
        if(check(mid))  high=mid;
        else  low=mid+1;
    }
    return low;
}
int main()
{
    while(scanf("%d%d%d",&k,&c,&m)!=EOF)
    {
        for(int i=1; i<=k+c; i++)
            for(int j=1; j<=k+c; j++)
            {
                scanf("%d",&mat[i][j]);
                if(mat[i][j]==0)  mat[i][j]=INF;
            }
        floyed();
        printf("%d/n",find());
    }
    return 0;
}