Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 51847 Accepted Submission(s): 21829

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

Author

Teddy

Source

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

Recommend

lcy | We have carefully selected several similar problems for you: 1203 2159 2955 1171 2191

状态转移方程：dp[i][j] = max{dp[i+1][j], dp[i+1][j-vol[i]]+val[i]};其中i表示第i个物品，j表示背包当前的容量，而数组中所存放的元素为在当前状态下的最大价值。通过下面的每个状态的最大值向上推。因此，最终的最大状态应该存放在dp[0][v]中。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 1000 + 10;
const int INF = 0x3f3f3f3f;
int  n, v, val[MAXN], vol[MAXN], dp[MAXN][MAXN];
int main() {int t; scanf("%d", &t);while (t--) {scanf("%d %d", &n, &v);for (int i = 0; i < n; i++) {scanf("%d", &val[i]);}for (int i = 0; i < n; i++) {scanf("%d", &vol[i]);}memset(dp, 0, sizeof(dp));for (int i = n - 1; i >= 0; i--) {for (int j = 0; j <= v; j++) {if (j < vol[i]) dp[i][j] = dp[i + 1][j];else dp[i][j] = max(dp[i+1][j], dp[i+1][j-vol[i]]+val[i]);}}printf("%d\n", dp[0][v]);}return 0;
}

转载于:https://www.cnblogs.com/cniwoq/p/6770830.html

HDU Problem 2062 Bone Collector【01背包】相关推荐

hdu 2602 Bone Collector 01背包
Bone Collector Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) T ...
【HDU - 2639】Bone Collector II （第K大背包，dp，STLset）
题干: The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup ...
hdu 6149 Valley Numer II(01背包套状压dp)
题目链接:hdu 6149 Valley Numer II 题意: 给你N个点,有k个为高点,其他为低点,现在这N个点有m条边,问你最多能组成多少个两个高点一个低点,低点和两个高点都有边相连这样的状态 ...
hdu 1864 最大报销额01背包dp
最大报销额 Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other) Total Subm ...
hdu 2546 饭卡（01背包）
题意:只要最后留下的钱够5元,就可以买下最贵的食物,使得余额最低,其他的钱的使用就是一个最大背包问题了. 代码: #include<iostream> #include<cstrin ...
HDu 3449 (有依赖的01背包) Consumer
题意: 有n件物品,对应有不同的价格和价值,这是典型的01背包.但现在有了一个限制,要买物品先买能装这件物品的特定的盒子,盒子的价值为0 代码理解得还不是太好,感觉这是一个"二重" ...
hdu 6092 Rikka with Subset 01背包思维
dp[i][j]表示前i个元素,子集和为j的个数.d[i][j] = d[i][j] + d[i-1][j-k] (第i个元素的值为k).这里可以优化成一维数组比如序列为 1 2 3,每一步的dp值 ...
HDU Bone Collector (01背包问题)
之前背包问题学了忘学了忘= =是太笨了. 题目许多年前,在泰迪的家乡,有一个人被称为"骨收集者".这个人喜欢收集各种骨头,例如狗,牛的骨头,他也去了坟墓-- 骨头收集者有一个大袋 ...
hdu 2546 饭卡【01背包】
题目链接:https://vjudge.net/contest/103424#problem/C 饭卡 Time Limit: 5000/ ...

HDU Problem 2062 Bone Collector【01背包】

Bone Collector

HDU Problem 2062 Bone Collector【01背包】相关推荐

最新文章

热门文章