hdoj 2602(背包)】的更多相关文章

意甲冠军:给出的数量和袋骨骼的数,然后给每块骨骼的价格值和音量.寻求袋最多可容纳骨骼价格值 难度;这个问题是最基本的01背包称号,不知道的话,推荐看<背包9说话> AC by SWS 主题链接 http://acm.hdu.edu.cn/showproblem.php?pid=2602 代码: #include<stdio.h> #include<string.h> typedef struct{ int w, v; }str; str s[1005]; int dp[…
Problem D Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 9   Accepted Submission(s) : 6 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Many years ago , in Teddy’s h…
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…
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 volu…
本文文旨,如题... 转载请注明出处... HDOJ 1176 免费馅饼 http://acm.hdu.edu.cn/showproblem.php?pid=1176 类似数塔,从底往上推,每次都是从下面三个中选最大的 #include<cstdio> #include<cstring> #define MAXN 100005 ];//第i秒第j个位置的馅饼数 int max1(int a,int b) { return a>b?a:b; } int max2(int a,i…
HDOJ(HDU).2602 Bone Collector (DP 01背包) 题意分析 01背包的裸题 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 1005 using namespace std; int v[nmax],w[nmax],dp[nmax]; int main() { //freopen("in…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 Bone Collector Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 54132    Accepted Submission(s): 22670 Problem Description Many years ago , in…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 题目分析:0-1背包  注意dp数组的清空, 二维转化为一维后的公式变化 /*Bone Collector Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 34192 Accepted Submission(s): 14066 Proble…
题目链接:HDOJ - 1171 题目大意 有 n 种物品,每种物品有一个大小和数量.要求将所有的物品分成两部分,使两部分的总大小尽量接近. 题目分析 令 Sum 为所有物品的大小总和.那么就是用给定的物品做完全背包,背包容量为 (Sum / 2) ,得到的结果是较小的一部分的大小. 完全背包问题可以使用单调队列优化,O(nm) . 代码 #include <iostream> #include <cstdio> #include <cstdlib> #include…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 思路分析:该问题为经典的0-1背包问题:假设状态dp[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值,则可以推导出dp递推公式 dp[i][v] = Max{dp[i-1][v], dp[i-1][v – c[i]] + w[i]}:c[i]表示第i件物品的容量,w[i]表示第i件物品的价值:该动态规划问题每个阶段的决策为是否要 选择第i件物品放入背包中,如果不选择第i件物…