poj 2392 Space Elevator

Space Elevator

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8893 Accepted: 4219

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.

排序+多重背包

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <set>#include <queue>using namespace std;#define maxn 40000+10#define inf 0x7ffffffpair <int ,int > p;int dp[maxn];int used[maxn];int n;struct Node{    int h;    int a;    int c;};Node arr[400+5];//int cmpD(Node a,Node b){//    return a.h > b.h;//}int cmpI(Node a,Node b){    return a.a < b.a;}int main(){    while(scanf("%d",&n) != EOF){        for(int i = 0; i < n; i++){            scanf("%d%d%d",&arr[i].h,&arr[i].a,&arr[i].c);        }        sort(arr,arr+n,cmpI);//        for(int i = 0; i < n; i++){//            printf("%d ",arr[i].a);//        }//        printf("\n");        memset(dp,0,sizeof(dp));        dp[0] = 1;        for(int i = 0; i < n; i++){            memset(used,0,sizeof(used));            for(int j = arr[i].h; j <= arr[i].a;j++){                if(dp[j-arr[i].h] && used[j-arr[i].h] < arr[i].c && !dp[j]){                    dp[j] = 1;                    used[j] = used[j-arr[i].h] +1;                }            }        }//        for(int i = 0; i < n; i++){//            printf("%d ",arr[i].a);//        }//        printf("\n");        for(int i = arr[n-1].a; i >= 0; i--){            if(dp[i]){                printf("%d\n",i);                break;            }        }    }    return 0;}