njupt-1134-Christmas

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Christmas

时间限制(普通/Java) : 2000 MS/ 6000 MS 运行内存限制 : 65536 KByte
总提交 : 33 测试通过 : 11

比赛描述

Christmas day is coming . There will be a ball in Christmas Eve . N men and N women will take part in the ball . One man and one woman form a pair . As we know , if there is a great difference in age or in height , the pair will be very disappointed . Now we define the disappointment between one man and one woman as :

Where Hi is the height of person i and AGEi is the age of that person . Your task is to find a plan to form N pairs , with the max disappointment value minimized .

输入

Input may contain several test data sets.
For each data set , the first line contains an integer N ( 0 < N <= 500 ) ;
The first line is followed by 2N lines . The first N lines describe N men , and the last N lines describe N women . Each line contains two integers , the height and the age of a person .
You can assume that the height of one person is between 100cm and 200cm and the age between 10 and 60 .
Input is ended by N = 0 , which should not be processed.

输出

For each data set , print the answer in a single line.

样例输入

2
141 27
134 10
169 34
178 18
0

样例输出

1801

题目来源

NUPT ACM

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std ; const int Max = 99999999 ; int K,N,M ;             //二分图中x和y中点的数目int H[505],A[505] , Hw[505],Aw[505] ;bool visit[505] ;        //记录y中节点是否使用int result[505] ;      //记录当前与y节点相连的x的节点int map[505][505] ;    //记录连接x和y的边int low , high ,mid , way ; bool hungary(int u)                                        //匈牙利算法{    for (int v = 0; v < M; v ++)    {        if (!visit[v] && map[u][v] <= mid)           //map[u][v] <= mid，这句是二分的关键，在普通的二分图匹配map[i][j] = 0/1        {            visit[v] = true;             if (result[v]  == -1 || hungary(result[v]))            {                result[v] = u;                 return true;            }        }    }        return false;} int match(){    int ret;     memset(result, -1, sizeof(result));     ret = 0;     for (int u = 0; u < N; u ++)    {        memset(visit, false, sizeof(visit));         if (hungary(u))        {            ret ++;        }    }     return ret;} int main(){    while(scanf("%d",&K),K!=0)    {        memset(map,0,sizeof(map)) ;        N = K  ;        for(int i = 0 ;  i < N ;i ++)        {            scanf("%d %d",&H[i],&A[i]) ;        }         M = K ;        for(int j = 0 ; j < M ; j ++)        {           scanf("%d %d",&Hw[j],&Aw[j]) ;        }         for(int i = 0 ; i < N ; i ++)        {            for(int j = 0 ; j < M ; j ++)            {                map[i][j] = (H[i]-Hw[j])*(H[i]-Hw[j])+(A[i]-Aw[j])*(A[i]-Aw[j]) ;            }        }         low =  Max ;        high = - Max ;        for(int i = 0 ; i < N ; i ++)   //求出最大和最小的map[][];        {            for(int j = 0 ; j < M ; j ++)            {                low = min(low,map[i][j]) ;                high = max(high,map[i][j]) ;            }        }        while(low < high)                //二分过程        {            mid = (low + high ) / 2 ;            way = match() ;            if(way >= M)            {                high = mid ;            }            else             {                low = mid + 1 ;            }        }        printf("%d\n",low) ;    }}

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