POJ 2392 Space Elevator

Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6236		Accepted: 2883

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

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

 

多重背包问题，先按最大高度排好序，再背包，注意判断是完全背包还是01背包。

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int dp[40100] ;
struct ddd{
    int h,a,c ;
}f[410] ;
bool cmp(ddd a,ddd b){
    return a.a < b.a ;
}
int main(){
    int K;
    while(scanf("%d",&K)!=EOF){
        memset(dp,0,sizeof(dp)) ;
        int ans = 0;
        for(int i=1;i<=K;i++)
            scanf("%d%d%d",&f[i].h,&f[i].a,&f[i].c) ;
        sort(&f[1],&f[1]+K,cmp) ;
        for(int x=1;x<=K;x++){
            int a = f[x].a ,h = f[x].h ,c = f[x].c ;
            if(h * c >= a){
                for(int i=h;i<=a;i++)
                    if(dp[i-h]+h > dp[i])
                        dp[i] = dp[i-h] + h ;
            }    else{
                int temp = 1;
                while(temp < c){
                    for(int i=a;i>=h*temp;i--)
                        if(dp[i-temp*h] + temp * h > dp[i])
                            dp[i] = dp[i-temp*h] + temp * h;
                    c -= temp ;
                    temp <<= 1;
                }
                for(int i=a;i>=h*c;i--)
                    if(dp[i-c*h]+c*h>dp[i])
                        dp[i] = dp[i-c*h] + c * h;
            }    
        }
        for(int i=0;i<=f[K].a;i++)
            if(ans < dp[i])
                ans = dp[i];
        printf("%d\n",ans) ;
    }
    return 0;
}