poj2392 Space Elevator( dp ,背包)

奶牛们上天了。。 有n种材料，每种材料高hi ， 限制最大放置高度ai ， 有ci块， 问能搭多高

首先对每种材料按照ai排序， 然后从小到大， dp[i][j] 表示前i中材料在j的高度下，第i种能剩多少，最后只要从amax往下判断第一个>=0的就可以了

题目：

Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7638		Accepted: 3606

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.

Source

USACO 2005 March Gold

 

 

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
//#pragma comment (linker,"/STACK:1024000000,1024000000")
#define MAXN 500
#define INF 0x7ffffff
int n;
struct P
{
    int h,c,a;

    bool operator < (const P& x) const
    {
        return a<x.a;
    }
};

int dp[MAXN][40000+10];

P w[MAXN];
int hmax = 0;
void init()
{
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>w[i].h>>w[i].a>>w[i].c;
        hmax  = max(hmax , w[i].a);
    }
    sort(w,w+n);
    for(int i=0;i<=n;i++)
    {
        for(int ta=0;ta<=hmax;ta++)
            dp[i][ta] = -1;
    }
    dp[0][0] = 0;
}

int main()
{
    init();

    for(int i=1;i<=n;i++)
    {
        for(int ta = 0 ; ta<=w[i-1].a;ta++)
        {
            if( dp[i-1][ta] >=0 )
            {
                dp[i][ta]  = w[i-1].c;
            }
            else if( ta<w[i-1].h|| dp[i][ ta - w[i-1].h]<=0)
            {
                dp[i][ta] = -1;
            }
            else
            {
                dp[i][ta] = dp[i][ ta- w[i-1].h] - 1;
            }
        }
    }

    for(int i=hmax;i>=0;i--)
    {
        if(dp[n][i]>=0)
        {
            cout<<i<<endl;
            break;
        }
    }

    return 0;
}