HDU-2602-Bone Collector （最基础DP！！）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31922    Accepted Submission(s): 13138

Problem Description

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 2³¹).

 

Sample Input

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

Author

Teddy

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

相信大家都对DP很熟了...但是我不是很熟.....上次北京现场赛就考了好多DP....差点压制死:-(

先贴个代码纪念纪念：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int dp[1005][1005], val[1005], vol[1005];int main(){int n, v, T;scanf("%d", &T);while(T--){int i, j;memset(dp, 0, sizeof(dp));scanf("%d %d", &n, &v);for(i=1; i<=n; i++){scanf("%d", &val[i]);}for(i=1; i<=n; i++){scanf("%d", &vol[i]);}for(i=1; i<=n; i++){for(j=0; j<=v; j++){if(j >= vol[i]) dp[i][j] = max(dp[i-1][j], dp[i-1][j-vol[i]]+val[i]);else dp[i][j] = dp[i-1][j];}}printf("%d\n", dp[n][v]);}return 0;}