专题三--1017

Posted 2020-07-13 Pic

题目

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

思考过程

这是一个01背包的模板题。
只要找出状态转移方程就解决了一大半问题。
这个问题我使用了一维数组进行了空间优化，因为题目给的数据的范围是N<=1000;V<=1000。所以需要开一个1000*1000的二位数组,这个数组太大了，超出了题目要求的范围，所以我使用了一个一位数组进行数组滚动。
状态转移方程为：

F[v]=max{ F[v-Ci]+Wi , F[v] }

需要特别注意的一个问题是，由于01背包问题需要比较的是上一个循环中F[v-Ci]+Wi , F[v]的值，因此需要递减循环F[v]
做题期间出现了一个小插曲，我写完之后提交三次，全部WA，全部绿，整的我脸都绿了。仔细检查代码后一点问题也没有，回头看题目才发现自己将题目中的数据顺序看错。。以后我一定要写一篇论文，名字就叫做《论ACM中英语与仔细读英语题目的重要性》。。

AC代码

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int v[1010];
int c[1010];
int dp[1010];
int MAX(int a, int b){
    return a > b ? a : b;
}
int main(){
    int T, N, V;
    cin >> T;
    while (T--){
        scanf("%d%d", &N, &V);
        for (int i = 1; i <= N; i++){
            scanf("%d", &c[i]);
        }
        for (int j = 1; j <= N; j++){
            scanf("%d", &v[j]);
        }
        for (int i = 0; i <= 1000; i++){
            dp[i] = 0;
        }
        for (int k = 1; k <= N; k++){
            for (int l = V; l >=v[k]; l--){
                dp[l] = MAX(dp[(l - v[k])] + c[k], dp[l]);
            }
        }
        printf("%d\n", dp[V]);
    }
}

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