POJ 2392 Space Elevator （多重背包 + 思路题）

题目链接：http://poj.org/problem?id=2392

Space Elevator

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7783 Accepted: 3690

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：给你k种blocks，高度为hi，每种有ci个，约束条件是 想放置这个blocks的时候，当前总高度不超过ai。

分析：思考，如何将石块排序，能保证做01背包的正确性，有兴趣的可以看这里的第G题。

给代码具体去思考吧：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int n;class blocks{public:int h,l;bool operator <(const blocks &b)const{return b.h+l<b.l+h;}}a[4000];int dp[40005];int max(int a,int b){return a>b?a:b;}int min(int a,int b){return a<b?a:b;}void binary_div(int h,int limit,int num){int i = 1;while(num){if(num >= i){a[n].h = h*i;a[n++].l = limit;num -= i;}else{a[n].h = h*num;a[n++].l = limit;num = 0;}i <<= 1;}}int main(){int t,h,ai,num,m;while(scanf("%d",&t)!=EOF){m = 0;n = 0;for (int i = 0; i < t; i++){scanf("%d %d %d",&h,&ai,&num);binary_div(h,ai,num);m = max(m,ai);}memset(dp,0,sizeof(dp));sort(a,a+n);//排序int ans = 0;for (int i = 0; i < n; i++){for (int j = a[i].l; j >= a[i].h; j--){dp[j] = max(dp[j],dp[j-a[i].h]+a[i].h);ans = max(dp[j],ans);}}printf("%d\n",ans);}return 0;}