POJ 2392 Space Elevator 多重背包

看见啸爷再看这道题目，也跟着看了看，看懂之后感觉有戏，很裸的背包，果然1A，嘿嘿。

题目大意：给你n种木块，然后让你输出最高可以组成的高度。

限制条件是：每种木块的个数，与木块的在高度h以上就不可以再出现了。

解题思路：根据每种木块可以到达的高度sort一遍然后就是多重背包，找到满足条件的最大的高度。

注意可以到达的最大的高度不会超过sort之后木块可以到达的上限，f[n-1].lim，因为在往上，是不能再搭木块的啊，所以就把这个高度当作，dp的上限，改了一下竟然省了300+ms。。。

Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7657		Accepted: 3621

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

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
#define M 1000100
//#define LL __int64
#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898

const int maxn = 500100;
using namespace std;

int dp[maxn];
struct node
{
int h, num, lim;
} f[maxn];
int cmp(node a, node b)
{
return a.lim < b.lim;
}

int main()
{
int n;
cin >>n;
for(int i = 0; i < n; i++)
cin >>f[i].h>>f[i].lim>>f[i].num;
sort(f , f+n, cmp);
memset(dp , 0 , sizeof(dp));
dp[0] = 1;
for(int i = 0; i < n; i++)
{
for(int j = f[n-1].lim; j >= 0; j--)
{
if(!dp[j])
continue;
for(int k = 1; k <= f[i].num; k++)
{
if(j+k*f[i].h <= f[i].lim)
{
if(!dp[j+k*f[i].h])
dp[j+k*f[i].h] = 1;
else
break;
}
}
}
}
int i;
for(i = f[n-1].lim; i >= 0; i--)
if(dp[i])
break;
cout<<i<<endl;
return 0;
}