poj2392(Space Elevator)

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.

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;
}