poj——2392——Space Elevator（多重）

Description


The cows are going to space! They plan to achieve orbit by building a sort of space elevator: a giant tower of blocks. They have K (1 <= K <= 400) different types of blocks with which to build the tower. Each block of type i has height h_i (1 <= h_i <= 100) and is available in quantity c_i (1 <= c_i <= 10). Due to possible damage caused by cosmic rays, no part of a block of type i can exceed a maximum altitude a_i (1 <= a_i <= 40000). 


Help the cows build the tallest space elevator possible by stacking blocks on top of each other according to the rules.
Input


* Line 1: A single integer, K 


* Lines 2..K+1: Each line contains three space-separated integers: h_i, a_i, and c_i. Line i+1 describes block type i.
Output


* Line 1: A single integer H, the maximum height of a tower that can be built
Sample Input
有一群奶牛想到太空去，他们有k中类型的石头，每一类石头高h，石头能达到的高度c，以及它的数量a，在做背包前需要对石块能到达的最大高度（a）进行排序，而且每种砖块都有一个限制条件，就是说以该种砖块结束的最大高度H不能超过某个高度，不同砖块的高度不同。求最高的高度是多少。
3
7 40 3
5 23 8
2 52 6
Sample Output

48

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>using namespace std;const int M=405;const int MAX=40005;struct node{int h;int a;int c;}Node[M];int cmp(node a,node b){return a.a<b.a;}int dp[MAX];int vis[MAX];int main(){int t;while(scanf("%d",&t)!=EOF){for(int i=1;i<=t;i++)scanf("%d%d%d",&Node[i].h,&Node[i].a,&Node[i].c);sort(Node+1,Node+t+1,cmp);memset(vis,0,sizeof(vis));vis[0]=1;int max=0;for(int i=1;i<=t;i++){memset(dp,0,sizeof(dp));for(int j=Node[i].h;j<=Node[i].a;j++){if(!vis[j]&&vis[j-Node[i].h]&&dp[j-Node[i].h]+1<=Node[i].c){    vis[j]=1;    dp[j]=dp[j-Node[i].h]+1;if(j>max)max=j;}}}printf("%d\n",max);}return 0;}