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.
题意：给你一些石头，告诉你每种石头的高度、能堆的最大高度、数量，，求所有石头能堆最大高度是多少。
题解：先按能堆的最大高度对石头从小到大进行排序，然后就是一个多重背包问题。
#include <iostream>#include <sstream>#include <fstream>#include <string>#include <map>#include <vector>#include <list>#include <set>#include <stack>#include <queue>#include <deque>#include <algorithm>#include <functional>#include <iomanip>#include <limits>#include <new>#include <utility>#include <iterator>#include <cstdio>#include <cstdlib>#include <cstring>#include <cctype>#include <cmath>#include <ctime>using namespace std;struct block{int h, a, c;bool operator < (const block& b) const{return a < b.a;}};int k, sum[40010], dp[40010];block s[410];int main(){cin >> k;for (int i = 0; i < k; ++i)scanf("%d%d%d", &s[i].h, &s[i].a, &s[i].c);sort(s, s+k);dp[0] = 1;int ans = 0;for (int i = 0; i < k; ++i){memset(sum, 0, sizeof(sum));for (int j = s[i].h; j <= s[i].a; ++j)if (dp[j-s[i].h] && !dp[j] && sum[j-s[i].h] < s[i].c){dp[j] = 1;sum[j] = sum[j-s[i].h] + 1;ans = max(ans, j);}}cout << ans << endl;return 0;}
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