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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 builtSample Input
3 7 40 3 5 23 8 2 52 6Sample Output
48Hint
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.
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 const int MAX = 40010; 6 const int INF = 0x3f3f3f3f; 7 8 struct Node 9 { 10 int h; 11 int a; 12 int c; 13 }; 14 15 int ans; 16 bool vis[MAX]; 17 Node block[MAX]; 18 19 bool cmp(Node a, Node b) 20 { 21 return (a.a < b.a); 22 } 23 24 void zero_pack(int w,int v) 25 { 26 for(int i = v; i >= w; i--) 27 if(vis[i-w]) 28 { 29 vis[i] = true; 30 ans = max(ans,i); 31 } 32 } 33 34 void complete_pack(int w, int v) 35 { 36 for(int i = w; i <= v; i++) 37 if(vis[i - w]) 38 { 39 vis[i] = true; 40 ans = max(ans,i); 41 } 42 } 43 44 void multi_pack(int w, int n, int v) 45 { 46 if(w * n > v) 47 complete_pack(w,v); 48 else 49 { 50 for(int i = 1; i < n; i *= 2) 51 { 52 n -= i; 53 zero_pack(w * i, v); 54 } 55 56 zero_pack(n * w,v); 57 } 58 } 59 60 int main() 61 { 62 #ifdef OFFLINE 63 freopen("in.txt", "r", stdin); 64 freopen("out.txt", "w", stdout); 65 #endif 66 67 ans = 0; 68 memset(vis,false,sizeof(vis)); 69 70 int k; 71 scanf("%d", &k); 72 for(int i = 0; i < k; i++) 73 scanf("%d%d%d", &block[i].h, &block[i].a, &block[i].c); 74 75 vis[0] = true; 76 sort(block, block + k, cmp); 77 78 for(int i = 0; i < k; i++) 79 multi_pack(block[i].h, block[i].c, block[i].a); 80 81 printf("%d\n", ans); 82 return 0; 83 84 }
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原文地址：http://www.cnblogs.com/gwsbhqt/p/4710156.html
