POJ 2392 Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11172		Accepted: 5317

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

题目大意：n 种 石头 有每块石头的高度 每块石头的能所处的最大高度  以及每块石头的块数

求 能到的最大高度

思路: 按照 每种石头能所 处 得最大高度从小到大排序 （有贪心的思想）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
struct node{
    int h;
    int a;
    int c;
    bool operator < (const node &xxx)const 
    {
        return a<xxx.a;
    }
}e[600]; 
int f[400000],num[400000];
int main()
{
    cin>>n;
    for(int i=0;i<n;i++)
     cin>>e[i].h>>e[i].a>>e[i].c;
    sort(e,e+n);
    memset(f,0,sizeof f );
    f[0]=1;
    int ans=0;
    for(int i=0;i<n;i++)
    {
        memset(num,0,sizeof num );
        for(int j=e[i].h;j<=e[i].a;j++)
        {
            if(!f[j]&&f[j-e[i].h]&&num[j-e[i].h]<e[i].c)// !f[j]  表示f[j] 没遍历 到过 即 非最优解
            {   //  f[j-e[i].h]为真 所以保证 f[j]能走到
                f[j]=1;  
                num[j]=num[j-e[i].h]+1;//统计该石块放置的个数  
                if(ans<j)  
                  ans=j;
            }
        }
    }
    
    
    printf("%d",ans);
    return 0;
}