題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=3392

Pie

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 793    Accepted Submission(s): 214


Problem Description A lot of boys and girls come to our company to pie friends. After we get their information, we need give each of them an advice for help. We know everyone’s height, and we believe that the less difference of a girl and a boy has, the better it is. We need to find as more matches as possible, but the total difference of the matches must be minimum.
Input The input consists of multiple test cases. The first line of each test case contains two integers, n, m (0 < n, m <= 10000), which are the number of boys and the number of girls. The next line contains n float numbers, indicating the height of each boy. The last line of each test case contains m float numbers, indicating the height of each girl. You can assume that |n – m| <= 100 because we believe that there is no need to do with that if |n – m| > 100. All of the values of the height are between 1.5 and 2.0.
The last case is followed by a single line containing two zeros, which means the end of the input.
Output Output the minimum total difference of the height. Please take it with six fractional digits.
Sample Input 2 3 1.5 2.0 1.5 1.7 2.0 0 0
Sample Output 0.000000
Author

[email protected]
Source
思路：dp+滾動陣列

（1）：要求最佳匹配，首先得將兩陣列從小到大排序~

 （2）： 然後再明確dp[][]表示的意思;dp[i][j] 表示a陣列中前i個數和b陣列中前j個數匹配的最優解

  （3）：接下來 看看狀態轉移方程； if(i==j)dp[i][j]=dp[i-1][j-1]+fabs(a[i]-b[j]);

                                                                   else dp[i][j]=min(dp[i-1][j-1]+fabs(a[i]-b[i]),dp[i][j-1]);

   (4):   因為題目中的n最大取到10000，如果開個陣列dp[10000][10000],那麼執行不了~那麼再觀察觀察狀態轉移方程，發現當前這個數是由它左邊這列遞推過來的，我們可以用一個dp[2][10000]的滾動陣列即可，因為我只關心最後一個dp[n][m]值，所以前面的一些值被覆蓋不影響我後面的求值過程；（可以在紙上畫一畫，就知道這個滾動陣列了）

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <cstdio>
#include <algorithm>
#include <cmath>
const int maxn=11000;
using namespace std;

double a[maxn],b[maxn];
double dp[2][maxn];

int main()
{
 int n,m;
 while(cin>>n>>m)
 {
        if(n==0&&m==0)break;
        for(int i=1;i<=n;i++)scanf("%lf",&a[i]);
        for(int i=1;i<=m;i++)scanf("%lf",&b[i]);

        double *A=a,*B=b;
        if(n>m){swap(n,m);swap(A,B);}
        sort(A+1,A+1+n);
        sort(B+1,B+1+m);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
         for(int j=i;j<=i+m-n;j++)
         {
          if(i==j)
          {
              dp[i&1][j]=dp[(i-1)&1][j-1]+fabs(A[i]-B[j]);
              //printf("dp[%d][%d] :%.6lf",i,j,dp[i&1][j]);
          }
          else
          {
              dp[i&1][j]=min(dp[(i-1)&1][j-1]+fabs(A[i]-B[j]),dp[i&1][j-1]);
              //printf("dp[%d][%d] :%.6lf",i,j,dp[i&1][j]);
          }
         }
       printf("%.6lf\n",dp[n&1][m]);
 }
 return 0;
}