POJ 2392 Space Elevator

Space Elevator

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8778 Accepted: 4168

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：

        给出k段高度为h，数量为c，并且高度不超过c的木块，求建造的最高的塔

题解：

        肯定要先根据木块的最高高度c排序,然后用背包来解决。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=40000+100;int dp[maxn];struct node{    int h;    int a;    int c;}p[505];bool cmp(node x,node y){    return x.a<y.a;}int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)        {            scanf("%d%d%d",&p[i].h,&p[i].a,&p[i].c);        }        sort(p,p+n,cmp);        memset(dp,0,sizeof(dp));        dp[0]=1;        int ans=0;        for(int i=0;i<n;i++)        {            for(int j=p[i].a;j>0;j--)//最高高度开始，不让前面的操作影响后面的            {                for(int k=p[i].c;k>=0;k--)//使用的第i种木块不断增多，还可以用个count数组降低复杂度，此循环不要                {                    if(j-p[i].h*k>=0&&dp[j-p[i].h*k])                    {                        ans=max(ans,j);                        dp[j]=1;                    }                }            }        }        printf("%d\n",ans);    }    return 0;}