POJ 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.

题目大意

有一群牛要上太空,他们计划建一个太空梯(用一些石头垒),他们有k种不同类型的石头,每一种石头的高度为h,数量为c,由于会受到太空辐射,每一种石头不能超过这种石头的最大建造高度a,求解利用这些石头所能修建的太空梯的最高的高度.

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.

题解

记忆化搜索+贪心。把两个数组写成bool型的了。哭瞎了，调了半个小时。

#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#include<cstdlib>#include<algorithm>using namespace std;int n,ans;struct shi{int h,a,c;} s[402];int pd[40002],usd[40002];bool kp(const shi &x,const shi &y){return x.a<y.a;}int main(){scanf("%d",&n);for(int i=1;i<=n;i++)    scanf("%d%d%d",&s[i].h,&s[i].a,&s[i].c);sort(s+1,s+n+1,kp);memset(pd,0,sizeof(pd));pd[0]=1; ans=0;for(int i=1;i<=n;i++)   {memset(usd,0,sizeof(usd));for(int j=s[i].h;j<=s[i].a;j++)   {if((!pd[j]) && pd[j-s[i].h] && usd[j-s[i].h]+1<=s[i].c)   {pd[j]=1;usd[j]=usd[j-s[i].h]+1;if(j>ans) ans=j;   }   }   }printf("%d\n",ans);return 0;}