poj2392

Space Elevator

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 12123 Accepted: 5769

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

题意：有n种石头，每个石头的高度hi，允许用的最高高度为ai，数量为ci，求能组合的最高高度。

思路：先给这些石头排个序，最高高度小的在前面，这样才能使得高度最大，然后再按多重背包来做，这样就可以了。

代码：

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;struct node{    int h,a,c;}a[405];bool cmp(node a,node b){    return a.a<b.a;}int dp[40005];int main(){    int k;    while(cin>>k)    {        int i,j,z;        for(i=0;i<k;i++)        {            scanf("%d %d %d",&a[i].h,&a[i].a,&a[i].c);        }        sort(a,a+k,cmp);        memset(dp,0,sizeof(dp));        dp[0]=1;        for(i=0;i<k;i++)        {            for(j=0;j<a[i].c;j++)            {                for(z=a[i].a;z>=a[i].h;z--)                {                    dp[z]|=dp[z-a[i].h];                }            }        }        for(i=a[k-1].a;i>=0;i--)        {            if(dp[i])            {                cout<<i<<endl;                break;            }        }    }    return 0;}