hdu 3127 WHUgirls（完全背包）

WHUgirls

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 2251    Accepted Submission(s): 852

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

12 4 42 2 23 3 9

 

Sample Output

9

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3127

题目大意：已知长宽分别为X,Y的大矩形，以及n个长和宽固定的小矩形的权值。求大矩形分割成小矩形后的能获得的最大权值。

解题思路：完全背包问题，dp[i][j]记录长宽为i,j的矩形分割后能获得的最大权值。

  递推方程为： dp[i][j]=max(dp[i][j],dp[i-r[k].x][j]+dp[r[k].x][j-r[k].y]+r[k].v);

                    dp[i][j]=max(dp[i][j],dp[i-r[k].x][r[k].y]+dp[i][j-r[k].y]+r[k].v);      

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int const maxn=1005;int dp[maxn][maxn];struct rec{    int x,y,v;}r[15];int main(){    int t,n,X,Y,i,j,k;    scanf("%d",&t);    while(t--)    {        scanf("%d %d %d",&n,&X,&Y);        memset(dp,0,sizeof(dp));        for(i=0;i<n;i++)            scanf("%d%d%d",&r[i].x,&r[i].y,&r[i].v);        for(i=1;i<=X;i++)            for(j=1;j<=Y;j++)                for(k=0;k<n;k++)                {                    if(r[k].x<=i&&r[k].y<=j)                    {                        dp[i][j]=max(dp[i][j],dp[i-r[k].x][j]+dp[r[k].x][j-r[k].y]+r[k].v);                        dp[i][j]=max(dp[i][j],dp[i-r[k].x][r[k].y]+dp[i][j-r[k].y]+r[k].v);                    }                    if(r[k].y<=i&&r[k].x<=j)                    {                        dp[i][j]=max(dp[i][j],dp[i-r[k].y][j]+dp[r[k].y][j-r[k].x]+r[k].v);                        dp[i][j]=max(dp[i][j],dp[i-r[k].y][r[k].x]+dp[i][j-r[k].x]+r[k].v);                    }                }        printf("%d\n", dp[X][Y]);    }    return 0;}