Space Elevator
Time Limit: 1000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 239264-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.
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.
* 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.
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。。。
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 }