POJ2392-Space Elevator

Space Elevator

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11818 Accepted: 5601

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：一群牛要上太空，给出n种石块，每种石块给出单块高度，总高度不能超过的最大值，数量，用这些石块能组成的最大高度

解题思路：先进行一次排序，将最大高度小的放在前面，只有这样才能得到最优解，如果将大的放在前面，后面有的小的就不能取到，然后进行完全背包

#include <iostream>  #include <cstdio>  #include <cstring>  #include <string>  #include <algorithm>  #include <cmath>  #include <map>  #include <cmath>  #include <set>  #include <stack>  #include <queue>  #include <vector>  #include <bitset>  #include <functional>  using namespace std;#define LL long long  const int INF = 0x3f3f3f3f;struct node{int h, a, c;}x[409];int cmp(node x, node y){return x.a<y.a;}int dp[50090], sum[50090];int main(){int n;while (~scanf("%d", &n)){for (int i = 1; i <= n; i++) scanf("%d%d%d", &x[i].h, &x[i].a, &x[i].c);sort(x + 1, x + n + 1, cmp);memset(dp, 0, sizeof dp);dp[0] = 1;int ma = 0;for (int i = 1; i <= n; i++){memset(sum, 0, sizeof sum);for (int j = x[i].h; j <= x[i].a; j++){if (!dp[j] && dp[j - x[i].h] && sum[j - x[i].h] + 1 <= x[i].c){dp[j] = 1;sum[j] = sum[j - x[i].h] + 1;ma=max(ma, j);}}}printf("%d\n", ma);}return 0;}