Think 很明显的01背包问题。。。
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).
Example Input
1 5 10 1 2 3 4 5 5 4 3 2 1
Example Output 14
题目大意 ： 1.第一行输入T – T组数据 2.第二行N， M – 有N种骨头 ， 背包最大容量为M 3.每根骨头的 价值 4.每根骨头的体积 5.求 背包 的 最大价值 6.01背包
#include<bits/stdc++.h>using namespace std;int main() { int T; int v[1050]; int w[1050]; int dp[1050]; int i, j; int N, V; cin >> T; while(T --) { cin >> N >> V; for (i = 0;i <= N - 1; i ++) { cin >> v[i]; } for (i = 0;i <= N - 1; i ++) { cin >> w[i]; } memset(dp, 0, sizeof(dp)); for (i = 0;i <= N - 1;i ++) { for (j = V; j >= w[i];j --) { dp[j] = max(dp[j], dp[j - w[i]] + v[i]); } } cout << dp[V] <<endl; } return 0; }/***************************************************User name: Result: AcceptedTake time: 4msTake Memory: 172KBSubmit time: 2017-02-18 16:05:04****************************************************/新闻热点
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