HDU2602 Bone Collector（01背包）

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …

The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input The first line contain a integer T , the number of cases. Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.   Output One integer per line representing the maximum of the total value (this number will be less than 2 ³¹).   Sample Input 1 5 10 1 2 3 4 5 5 4 3 2 1   Sample Output 14

模板的01背包题

#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; struct Node { int h; int v; } node[1005]; int main() { int t,n,m,l; int dp[1005]; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); int i; for(i = 1; i<=n; i++) scanf("%d",&node[i].h); for(i = 1; i<=n; i++) scanf("%d",&node[i].v); memset(dp,0,sizeof(dp)); for(i = 1; i<=n; i++) { for(l = m; l>=node[i].v; l--) dp[l] = max(dp[l],dp[l-node[i].v]+node[i].h); } printf("%d\n",dp[m]); } return 0; }