Forming Quiz Teams （状压dp，最优配对问题）

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 betweenthehousesofgroup2andsoon. Youhavetomakesurethesummation(x1+x2+x3+...+xn) is minimized.

Input

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

题意：给你几个点（2*n）的坐标，现在想把这些点分为n对，使得n对的距离和最小。

题解：状压dp，dp[S]表示状态S里面所有点互相配对的最小值，状态转移方程：dp[S] = min{dis[i][j] + dp[S^(1<<i)^(1<<j)}，i表示状态S里面的最大点，j表示状态S里面小于i的点。解释一下，因为i必须要一个小于它的匹配点，所以dp[S]就该等于取出i，j点的状态dp[S']加上dis[i][j]，由此也可知道，当状态S其中的点为奇数的时候，dp[S]就不存在。

//#include"bits/stdc++.h"
//#include<unordered_map>
//#include<unordered_set>
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<set>
#include<vector>
#include<bitset>
#include<climits>
#include<queue>
#include<iomanip>
#include<cmath>
#include<stack>
#include<map>
#include<ctime>
#include<new>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define MT(a,b) memset(a,b,sizeof(a))
#define lson l, mid, node << 1
#define rson mid + 1, r, node << 1 | 1
const int INF  =  0x3f3f3f3f;
const int O    =  1e5;
const int mod  =  1e4 + 7;
const int maxn =  1e6+5;
const double PI  =  acos(-1.0);
const double E   =  2.718281828459;

double dp[1<<21];
int main(){
    int n, l = 0;
    while(~scanf("%d", &n) && n){
        n *= 2;
        double x[100], y[100];
        for(int i=0; i<n; i++) {
            char s[100]; scanf("%s %lf%lf", s, &x[i], &y[i]);
        }
        
        double dis[100][100];
        for(int i=0; i<n; i++) for(int j=0; j<n; j++)
            dis[i][j] = sqrt((x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]));
        
        dp[0] = 0;
        for(int S=1; S<(1<<n); S++){
            dp[S] = INF;
            int i =0, cnt = 0;
            for(int j=0; j<n; j++) if(S & (1<<j)) { i = j; cnt ++; }
            if(cnt & 1) continue;
            for(int j=0; j<i; j++) if(S & (1<<j))
                dp[S] = min(dp[S], dis[i][j] + dp[S^(1<<i)^(1<<j)]);
        }
        double ans = dp[(1<<n)-1];
        printf("Case %d: %.2f\n", ++l, ans);
    }
    return 0;
}

Forming Quiz Teams （状压dp，最优配对问题）

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