Bone Collector HDU - 2602（01背包题解报告）

冷不防 2021-07-24 13:36 274阅读 0赞

Bone Collector  
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  
1 5  
10 1 2 3 4 5 5 4 3 2 1

Sample Output  
14

#define _CRT_SECURE_NO_WARNINGS
    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    using namespace std;
    const int maxn = 1e5 + 10;
    int w[maxn], c[maxn],dp[maxn];/w[i]表示骨头的体积，c[i]表示骨头的价值
    int main()
    { 
    	int T;
    	scanf("%d", &T);
    	while (T--)
    	{ 
    		memset(dp, 0, sizeof(dp));
    		int n, V;
    		scanf("%d%d", &n, &V);
    		for (int i = 1; i <= n; i++)
    		{ 
    			scanf("%d", &c[i]);
    		}
    		for (int i = 1; i <= n; i++)
    		{ 
    			scanf("%d", &w[i]);
    		}		
    		for (int i = 1; i <= n; i++)
    		{ 
    			for (int v = V; v >= w[i]; v--)
    				dp[v] = max(dp[v], dp[v - w[i]] + c[i]);状态转移方程
    		}
    		printf("%d\n", dp[V]);
    	}
    	return 0;
    }