Space Elevator(多重揹包)
阿新 • • 發佈:2019-02-08
Problem 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
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
/*題意:有一群牛要上太空,他們計劃建一個太空梯(用一些石頭壘),他們有k種不同型別的石頭,每一種石頭的高度 為h,數量為c,由於會受到太空輻射,每一種石頭不能超過這種石頭的最大建造高度a,求解利用這些石頭所能修建的 太空梯的最高的高度.多重揹包問題,與一般的多重揹包問題所不同的知識多了一個限制條件就是某些"物品"疊加 起來的"高度"不能超過一個值,於是我們可以對他們的最高可能達到高度進行排序,然後就是一般的多重揹包問題了 */ #include <iostream> #include <cstring> #include <cstdio> #include <algorithm> using namespace std; int dp[60000],k[60000]; struct node { int h,c,a; }m[60000]; int n; int cmp(node x,node y) { return x.a<y.a; } int main() { while(cin>>n) { for(int i=1;i<=n;i++) { cin>>m[i].h>>m[i].a>>m[i].c; } sort(m+1,m+n+1,cmp); memset(dp,0,sizeof(dp)); dp[0]=1; int ans=0; for(int i=1;i<=n;i++) { memset(k,0,sizeof(k)); for(int j=m[i].h;j<=m[i].a;j++) { if(!dp[j]&&dp[j-m[i].h]&&m[i].c>k[j-m[i].h]) { dp[j]=1; k[j]=k[j-m[i].h]+1; if(j>ans)ans=j; } } } cout<<ans<<endl; } return 0; }