Forming Quiz Teams - UVa 10911 状压dp

最新推荐文章于 2019-07-06 17:59:27 发布

原创最新推荐文章于 2019-07-06 17:59:27 发布 · 481 阅读

0 ·

CC 4.0 BY-SA版权

dp动态规划和递推同时被 2 个专栏收录

281 篇文章

订阅专栏

UVa

218 篇文章

订阅专栏

本文介绍了一种基于状态压缩动态规划的方法，用于解决将学生两两分组以参加竞赛的问题，目标是最小化各组成员间的距离之和。

4^th IIUC Inter-University Programming Contest, 2005
G	Forming Quiz Teams
	Input: standard input Output: standard output
	Problemsetter: Sohel Hafiz

You have been given the job of forming the quiz teams for the next ‘MCA CPCI Quiz Championship’. There are2*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 member’s 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 file. 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 student’s name, followed by the xcoordinate 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	Output for 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	Case 1: 118.40 Case 2: 1.41

题意：每两个一组，使得每组的距离的和最小。

思路：典型的状压dp。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int num[20][40100];
double x[20],y[20],dp[140010],dis[20][20];
char s[1010];
int get1(int k)
{ int ans=0;
  while(k)
  { if(k&1)
     ans++;
    k/=2;
  }
  return ans;
}
int main()
{ int n,i,j,k,p,ret,t=0;
  k=1<<16;
  for(i=1;i<70000;i++)
   dp[i]=1000000000;
  for(i=0;i<k;i++)
  { p=get1(i);
    if(p%2==0)
    { num[p][0]++;
      num[p][num[p][0]]=i;
    }
  }
  while(~scanf("%d",&n) && n)
  { n*=2;
    for(i=1;i<=n;i++)
     scanf("%s%lf%lf",s,&x[i],&y[i]);
    for(i=1;i<70000;i++)
     dp[i]=1000000000;
    for(i=1;i<=n;i++)
     for(j=i+1;j<=n;j++)
     { dis[i][j]=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
       dis[j][i]=dis[i][j];
     }
    for(p=0;p<n;p+=2)
    {for(i=1;i<=n;i++)
      for(j=i+1;j<=n;j++)
       for(k=1;k<=num[p][0];k++)
       { if((num[p][k] & (1<<(i-1)))==0 && (num[p][k] & (1<<(j-1)))==0)
         { ret=num[p][k]+(1<<(i-1))+(1<<(j-1));
           dp[ret]=min(dp[ret],dp[num[p][k]]+dis[i][j]);
         }
       }
    }
    printf("Case %d: %.2f\n",++t,dp[(1<<n)-1]);
  }
}