http://poj.org/problem?id=2392
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.
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
题意:有一头奶牛要上太空,他有很多种石头,每种石头的高度是hi,
但是不能放到ai之上的高度,并且这种石头有ci个
将这些石头叠加起来,问能够达到的最高高度。
把多重背包转化为01背包
#include<cstdio>//把多重背包转化为01背包 #include<iostream> #include<cstring> #include<algorithm> using namespace std; struct DP { int h,a,c; }; DP num[400+10]; int dp[40000+100]; int cmp(DP x,DP y) { return x.a<y.a; } int main() { int n,i,j,k; while(scanf("%d",&n)!=EOF) { memset(dp,0,sizeof(dp)); dp[0]=1; for(i=1;i<=n;i++) { scanf("%d%d%d",&num[i].h,&num[i].a,&num[i].c); } sort(num+1,num+n+1,cmp); for(i=1;i<=n;i++) { for(j=1;j<=num[i].c;j++) { for(k=num[i].a;k>=0;k--) { if(dp[k]==1&&k+num[i].h<=num[i].a) { dp[k+num[i].h]=1; } } } } for(i=40000;i>=0;i--) { if(dp[i]==1) break; } printf("%d ",i); } return 0; }
二进制优化
#include <cstring> #include <cmath> #include <cstdio> #include <cstdlib> #include <algorithm> #include <queue> #include <iostream> using namespace std; int n,dp[45000]; struct node { int a,b,c; }e[500]; bool cmp(node a, node b) { return a.b < b.b; } int main() { int i,j,k; while(scanf("%d",&n)!=EOF) { for(i = 0; i < n ; i++) { scanf("%d %d %d",&e[i].a,&e[i].b,&e[i].c); } memset(dp, -1 , sizeof(dp)); sort(e,e + n, cmp); dp[0] = 1; for(i = 0; i < n; i++) { int sum = 1; int q; for( k = 1; k<=e[i].c; k *= 2) { q = k * e[i].a; for(j = e[i].b; j >=0; j --) { if(dp[j] != -1 && j + q <= e[i].b) { dp[j + q] = 1; } } e[i].c -= k; } q = e[i].c * e[i].a; for(j = e[i].b; j >=0; j --) { if(dp[j] != -1 && j + q <= e[i].b) { dp[j + q] = 1; } } } for(i = 40000; i >= 0; i--) { if(dp[i] == 1) break; } cout<<i<<endl; } return 0; }