Pie（求最小身高差，dp）

最新推荐文章于 2021-03-08 12:18:10 发布

转载最新推荐文章于 2021-03-08 12:18:10 发布 · 244 阅读

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#java #数据结构与算法

本文探讨了一个在给定男女身高情况下，通过排序和动态规划算法找到最小身高差异匹配的问题。详细介绍了输入输出格式，算法实现，以及如何通过滚动数组优化计算过程。

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Pie

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 895 Accepted Submission(s): 246

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

Sample Output

0.000000

题解：

输入n个男生，m个女生的身高。

把人数较少的一方和另外一方匹配完，求最少的差值。

分别把男女的身高排序。人数较少的一方的每个人可以匹配abs（m-n+1）个人。

因为|m-n|<=100 所以一个人最多匹配100人。

之后用二维滚动数组就可以了。

dp[i][j]男对女相错j个人的最小身高差；

dp[i%2][k]=min(dp[(i-1)%2][k]+abs(a[i]-b[j]),dp[i%2][k-1])

不知道为啥sort一直错qsort就对了。。。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define SD(x) scanf("%lf",&x)
#define P_ printf(" ")
typedef long long LL;
const int MAXN=10010;
double dp[2][110],a[MAXN],b[MAXN];
int cmp(const void *a,const void *b)
{
    return *(double *)a < *(double *)b ? 1 : -1;
}
int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m),n||m){
        if(n<=m){
            for(int i=1;i<=n;i++)SD(a[i]);
            for(int i=1;i<=m;i++)SD(b[i]);
        }
        else{
            swap(n,m);
            for(int i=1;i<=m;i++)SD(b[i]);
            for(int i=1;i<=n;i++)SD(a[i]);
        }
        qsort(a+1,n,sizeof(a[1]),cmp);
        qsort(b+1,m,sizeof(b[1]),cmp);
        int len=m-n+1;
        mem(dp,0);
        for(int i=1;i<=n;i++){
            dp[i%2][1]=dp[(i-1)%2][1]+abs(a[i]-b[i]);
            for(int k=2;k<=len;k++){
                int j=i+k-1;
                dp[i%2][k]=min(dp[(i-1)%2][k]+abs(a[i]-b[j]),dp[i%2][k-1]);
            }
        }
        printf("%.6lf\n",dp[n%2][len]);
    }
    return 0;
}