HDU3127：WHUgirls（二维完全背包）

最新推荐文章于 2025-08-08 22:01:10 发布

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最新推荐文章于 2025-08-08 22:01:10 发布

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文章标签： java

原文链接：http://www.cnblogs.com/junior19/p/6729979.html

本文介绍了一个布料切割问题，目标是最高效地将一块巨大的矩形布料切割成小矩形片段以制作围巾，并实现利润最大化。通过动态规划算法解决此问题，给出了完整的代码实现。

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WHUgirls

Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2810 Accepted Submission(s): 1052

Problem Description

There are many pretty girls in Wuhan University, and as we know, every girl loves pretty clothes, so do they. One day some of them got a huge rectangular cloth and they want to cut it into small rectangular pieces to make scarves. But different girls like different style, and they voted each style a price wrote down on a list. They have a machine which can cut one cloth into exactly two smaller rectangular pieces horizontally or vertically, and ask you to use this machine to cut the original huge cloth into pieces appeared in the list. Girls wish to get the highest profit from the small pieces after cutting, so you need to find out a best cutting strategy. You are free to make as many scarves of a given style as you wish, or none if desired. Of course, the girls do not require you to use all the cloth.

Input

The first line of input consists of an integer T, indicating the number of test cases.
The first line of each case consists of three integers N, X, Y, N indicating there are N kinds of rectangular that you can cut in and made to scarves; X, Y indicating the dimension of the original cloth. The next N lines, each line consists of two integers, xi, yi, ci, indicating the dimension and the price of the ith rectangular piece cloth you can cut in.

Output

Output the maximum sum of prices that you can get on a single line for each case.

Constrains
0 < T <= 20
0 <= N <= 10; 0 < X, Y <= 1000
0 < xi <= X; 0 < yi <= Y; 0 <= ci <= 1000

Sample Input

   1 2 4 4 2 2 2 3 3 9
  

Sample Output

9

Source

2009 Asia Wuhan Regional Contest Online

思路：每割下一个矩形就将剩余的布料分成两个矩形来计算即可。

#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;
int dp[1001][1001], a[1001], b[1001], c[1001];
int main()
{
    int t, n, x, y;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp, 0, sizeof(dp));
        scanf("%d%d%d",&n,&x,&y);
        for(int i=0; i<n; ++i)
            scanf("%d%d%d",&a[i],&b[i],&c[i]);
        for(int i=1; i<=x; ++i)
            for(int j=1; j<=y; ++j)
                for(int k=0; k<n; ++k)
                {
                    if(i>=a[k] && j>=b[k])
                        dp[i][j] = max(dp[i][j], max(dp[i-a[k]][j] + dp[a[k]][j-b[k]], dp[i][j-b[k]] + dp[i-a[k]][b[k]]) + c[k]);
                    if(i>=b[k] && j>=a[k])
                        dp[i][j] = max(dp[i][j], max(dp[i-b[k]][j] + dp[b[k]][j-a[k]], dp[i][j-a[k]] + dp[i-b[k]][a[k]]) + c[k]);
                }
        printf("%d\n",dp[x][y]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/junior19/p/6729979.html