HDU3127 二维完全背包 DP

WHUgirls

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 584    Accepted Submission(s): 262

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

 

 

二维的完全背包 .

设 dp[i][j] 为当布的尺寸为 i*j 的时候能够裁剪出的最大价值 ,xx[k] 和 yy[k] 为第 k 块布的尺寸 , 其状态转移方程如下 :

 

if(i>=xx[k]&&j>=yy[k]) dp[i][j]=max(dp[i][j],dp[i-xx[k]][yy[k]]+v[k]+dp[i][j-yy[k]],dp[xx[k]][j-yy[k]]+v[k]+dp[i-xx[k]][j]);

if (i>=yy[k]&&j>=xx[k])

dp[i][j]=max(dp[i][j],dp[i-yy[k]][xx[k]]+v[k]+dp[i][j-xx[k]],dp[yy[k]][j-xx[k]]+v[k]+dp[i-yy[k]][j]);

 

方程两段表示的是布横放和竖放的情况 .

代码如下 :

 

#include<stdio.h> #include<string.h> #define MAXN 20 int dp[1005][1005]; inline int max(int a,int b,int c) { if (a>b) return a>c?a:c; else return b>c?b:c; } inline int max(int a,int b) { return a>b?a:b; } int main() { int i,j,k,t,n,x,y,xx[MAXN],yy[MAXN],v[MAXN]; scanf("%d",&t); while (t--) { memset(dp,0,sizeof(dp)); scanf("%d%d%d",&n,&x,&y); for (i=1; i<=n; ++i) { scanf("%d%d%d",&xx[i],&yy[i],&v[i]); } for (i=1; i<=x; ++i) { for (j=1; j<=y; ++j) { for (k=1; k<=n; ++k) { if (i>=xx[k]&&j>=yy[k]) dp[i][j]=max(dp[i][j],dp[i-xx[k]][yy[k]]+v[k]+dp[i][j-yy[k]],dp[xx[k]][j-yy[k]]+v[k]+dp[i-xx[k]][j]); if (i>=yy[k]&&j>=xx[k]) dp[i][j]=max(dp[i][j],dp[i-yy[k]][xx[k]]+v[k]+dp[i][j-xx[k]],dp[yy[k]][j-xx[k]]+v[k]+dp[i-yy[k]][j]); } } } printf("%d/n",dp[x][y]); } return 0; }