poj2392 Space Elevator

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

3
7 40 3
5 23 8
2 52 6

Sample Output

48

这题先要对a_i进行升序排序，因为每一个长度都有最大的限度，所以从最大限度值最小的开始背包。然后注意初始值都设为-1，dp[0]=0,因为这里要求的是恰好到一个高度，并不是小于等于就行（有点抽象，仔细体会一下）。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int max(int a,int b){
	return a>b?a:b;
}
struct node{
	int h,a,num;
}b[500];
bool cmp(node c,node d){
	return c.a<d.a;
}
int dp[40050];

int main()
{
	int n,m,i,j,k,sum;
	while(scanf("%d",&n)!=EOF)
	{
		m=0;
		for(i=1;i<=n;i++){
			scanf("%d%d%d",&b[i].h,&b[i].a,&b[i].num);
			if(b[i].a>m)m=b[i].a;
		}
		sort(b+1,b+1+n,cmp);
		memset(dp,-1,sizeof(dp));
		dp[0]=0;
		for(i=1;i<=n;i++){
			if(b[i].h*b[i].num>=b[i].a){
				for(j=b[i].h;j<=b[i].a;j++){
					if(dp[j-b[i].h]!=-1)dp[j]=0;
				}
			}
			else{
				k=1;sum=0;
				while(sum+k<b[i].num){
					sum+=k;
					for(j=b[i].a;j>=k*b[i].h;j--){
						if(dp[j-k*b[i].h]!=-1)dp[j]=0;
					}
					k*=2;
				}
				k=b[i].num-sum;
				for(j=b[i].a;j>=k*b[i].h;j--){
					if(dp[j-k*b[i].h]!=-1)dp[j]=0;
				}
			}
		}
		for(i=m;i>=0;i--){
			if(dp[i]!=-1){
				printf("%d
",i);break;
			}
		}
	}
	return 0;
}