Space Elevator(DP)
2016-01-15 21:05
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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
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.
一堆砖头,有c个,高度h,只能在海拔a以下使用,问最高能搭多高。
一看范围- -直接把物品拆开写背包,非常暴力的写法,但是数据水,过了- -,其实应该是用之前一个方法,dp[i][j]表示i个砖头到达j高度剩下最多第i种砖头的个数,这个只是加了递推的条件,懒得再写一遍了。
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.
一堆砖头,有c个,高度h,只能在海拔a以下使用,问最高能搭多高。
一看范围- -直接把物品拆开写背包,非常暴力的写法,但是数据水,过了- -,其实应该是用之前一个方法,dp[i][j]表示i个砖头到达j高度剩下最多第i种砖头的个数,这个只是加了递推的条件,懒得再写一遍了。
#include<cstdio> #include<cstring> #include<iostream> #include<queue> #include<vector> #include<algorithm> #include<string> #include<cmath> #include<set> #include<map> #include<vector> #include<stack> #include<utility> #include<sstream> using namespace std; typedef long long ll; const int inf = 0x3f3f3f3f; const int maxn = 1005; int dp[40005]; struct Node { int h,a,c; }node[405]; bool cmp(Node a,Node b) { if(a.a == b.a) return a .h < b.h; return a.a < b.a; } int main() { #ifdef LOCAL freopen("C:\\Users\\ΡΡ\\Desktop\\in.txt","r",stdin); //freopen("C:\\Users\\ΡΡ\\Desktop\\out.txt","w",stdout); #endif // LOCAL int k; scanf("%d",&k); int he = 0; for(int i = 1;i <= k;i++) { scanf("%d%d%d",&node[i].h,&node[i].a,&node[i].c); he = max(he,node[i].a); } he = min(he,40000); memset(dp,0,sizeof(dp)); sort(node + 1,node + 1 + k,cmp); for(int i = 1;i <= k;i++) { dp[0] = 1; for(int j = 1;j <= node[i].c;j++) { for(int kk = node[i].a;kk >= node[i].h;kk--) dp[kk] = dp[kk] | dp[kk - node[i].h]; } } int ans = 0; for(int j = ans;j <= he;j++) if(dp[j])ans = max(ans,j); printf("%d\n",ans); return 0; }
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