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POJ 2392 Space Elevator

时间:2015-04-20 23:54:32      阅读:212      评论:0      收藏:0      [点我收藏+]

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Space Elevator

Time Limit: 1000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 2392
64-bit integer IO format: %lld      Java class name: Main
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

 
解题:多重背包吧。。。记得要排序,把最高高度低的放前面,因为低的放前面可以更好的更新后面的。。。
 
技术分享
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 const int maxn = 400010;
 7 struct node{
 8     int h,c,a;
 9     bool operator<(const node &t) const{
10         return a < t.a;
11     }
12 }b[401];
13 int dp[maxn],cnt[maxn];
14 int main(){
15     int n;
16     while(~scanf("%d",&n)){
17         for(int i = 0; i < n; ++i)
18             scanf("%d %d %d",&b[i].h,&b[i].a,&b[i].c);
19         sort(b,b+n);
20         memset(dp,0,sizeof dp);
21         int ans = 0;
22         for(int i = 0; i < n; ++i){
23             memset(cnt,0,sizeof cnt);
24             for(int j = b[i].h; j <= b[i].a; ++j){
25                 if(dp[j] < dp[j-b[i].h] + b[i].h && cnt[j-b[i].h] < b[i].c){
26                     dp[j] = dp[j-b[i].h] + b[i].h;
27                     cnt[j] = cnt[j-b[i].h] + 1;
28                     ans = max(ans,dp[j]);
29                 }
30             }
31         }
32         printf("%d",ans);
33     }
34     return 0;
35 }
View Code

 

POJ 2392 Space Elevator

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原文地址:http://www.cnblogs.com/crackpotisback/p/4442911.html

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