POJ 2392 Space Elevator

Space Elevator

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 2392
64-bit integer IO format: %lld      Java class name: Main

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

 

解题：多重背包吧。。。记得要排序，把最高高度低的放前面，因为低的放前面可以更好的更新后面的。。。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 400010;
struct node{
    int h,c,a;
    bool operator<(const node &t) const{
        return a < t.a;
    }
}b[401];
int dp[maxn],cnt[maxn];
int main(){
    int n;
    while(~scanf("%d",&n)){
        for(int i = 0; i < n; ++i)
            scanf("%d %d %d",&b[i].h,&b[i].a,&b[i].c);
        sort(b,b+n);
        memset(dp,0,sizeof dp);
        int ans = 0;
        for(int i = 0; i < n; ++i){
            memset(cnt,0,sizeof cnt);
            for(int j = b[i].h; j <= b[i].a; ++j){
                if(dp[j] < dp[j-b[i].h] + b[i].h && cnt[j-b[i].h] < b[i].c){
                    dp[j] = dp[j-b[i].h] + b[i].h;
                    cnt[j] = cnt[j-b[i].h] + 1;
                    ans = max(ans,dp[j]);
                }
            }
        }
        printf("%d",ans);
    }
    return 0;
}