poj 2392 Space Elevator (多重背包)

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

37 40 35 23 82 52 6

Sample Output

48

Hint

OUTPUT DETAILS: 

From the bottom: 3 blocks of type 2, below 3 of type 1, below 6 of type 3. Stacking 4 blocks of type 2 and 3 of type 1 is not legal, since the top of the last type 1 block would exceed height 40.

Source

USACO 2005 March Gold

题意：有一头奶牛要上太空，他有很多种石头，每种石头的高度是hi，但是不能放到ai之上的高度，并且这种石头有ci个
将这些石头叠加起来，问能够达到的最高高度

思路：多重背包，其次a要按从小到大排序；

      设ai<aj，如果aj先放，ai有可能不能将aj的值叠加,因为它所限制的值小一些；

代码：

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int N=40005;struct node{    int h,a,c;}str[N];int dp[N];bool cmp(node aa,node bb){    return aa.a<bb.a;}int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)            scanf("%d%d%d",&str[i].h,&str[i].a,&str[i].c);        memset(dp,0,sizeof(dp));        sort(str,str+n,cmp);        dp[0]=1;        int maxx=0;        for(int i=0;i<n;i++)        {            for(int j=str[i].a;j>=0;j--)            {                if(dp[j])                {                    for(int k=1;k<=str[i].c;k++)                    {                        int sum=j+k*str[i].h;                        if(sum>str[i].a) continue;                        dp[sum]=1;                        if(maxx<sum)                            maxx=sum;                    }                }            }        }        printf("%d\n",maxx);    }}