UVA_10911_FormingQuizTeams

10911 - Forming Quiz Teams

Time limit: 3.000 seconds

You have been given the job of forming the quiz teams for the next `MCA CPCI Quiz Championship'.

There are 2 N students interested to participate and you have to form N teams, each team consisting

of two members. Since the members have to practice together, all the students want their members

house as near as possible. Let x1 be the distance between the houses of group 1, x2 be the distance

between the houses of group 2 and so on. You have to make sure the summation (x1+x2+x3+: : :+xn)

is minimized.

Input

There will be many cases in the input le. Each case starts with an integer N (N  8). The next 2 N

lines will given the information of the students. Each line starts with the students name, followed by

the x coordinate and then the y coordinate. Both x; y are integers in the range 0 to 1000. Students

name will consist of lowercase letters only and the length will be at most 20.

Input is terminated by a case where N is equal to 0.

Output

For each case, output the case number followed by the summation of the distances, rounded to 2 decimal

places. Follow the sample for exact format.

Sample Input

5

sohel 10 10

mahmud 20 10

sanny 5 5

prince 1 1

per 120 3

mf 6 6

kugel 50 60

joey 3 24

limon 6 9

manzoor 0 0

1

derek 9 9

jimmy 10 10

0

Sample Output

Case 1: 118.40

Case 2: 1.41

这个问题可以使用状态压缩进一步提升程序效率

dfs直接也可以过

毕竟一次选两个最多16个，而且每个都要选并且没有顺序

16!!（隔一项的阶乘）还是可以接受的

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;

const int M=20;
double px[M];
double py[M];
double dis[M][M];   //点间距离
int isu[M];

double tot;

void dfs(int cn,int now,double t,int n)  //cn目前已经用了的数，now目前要填的位置，t目前的和，n总数
{
if(cn==n)
{
tot=min(tot,t);
return;
}
isu[now]=1;
for(int i=now+1;i<n;i++)            //这里注意用过没用过的关系
{
if(isu[i])
continue;
isu[i]=1;
int p=now+1;
while(isu[p])                  //寻找下个能填的位置
p++;
dfs(cn+2,p,t+dis[now][i],n);
isu[i]=0;
}
isu[now]=0;
}

int main()
{
int n;
//freopen("1.out","w",stdout);
int ca=1;
while(1)
{
scanf("%d",&n);
if(!n)
break;
tot=1e8;
memset(isu,0,sizeof(isu));
n*=2;
for(int i=0;i<n;i++)
scanf("%*s%lf%lf",&px[i],&py[i]);
for(int i=0;i<n;i++)
for(int j=i+1;j<n;j++)
dis[i][j]=sqrt((px[i]-px[j])*(px[i]-px[j])+(py[i]-py[j])*(py[i]-py[j]));
dfs(0,0,0,n);
printf("Case %d: %.2lf\n",ca++,tot); //注意输出格式
}
return 0;
}