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.
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
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
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; }