Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4824    Accepted Submission(s): 2514 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par…

/* 01背包第k优解问题 f[i][j][k] 前i个物品体积为j的第k优解 对于每次的ij状态 记下之前的两种状态 i-1 j-w[i] (选i) i-1 j (不选i) 分别k个 然后归并排序并且去重生成ij状态的前k优解 */ #include<iostream> #include<cstdio> #include<cstring> #define maxn 1010 using namespace std; ],x[maxn],y[maxn],a,b,z; i…

http://acm.hdu.edu.cn/showproblem.php?pid=2639       Problem Description The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it does…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4178    Accepted Submission(s): 2174 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took pa…

解题思路:对于01背包的状态转移方程式f[v]=max(f[v],f[v-c[i]+w[i]]);其实01背包记录了每一个装法的背包值,但是在01背包中我们通常求的是最优解, 即为取的是f[v],f[v-c[i]]+w[i]中的最大值,但是现在要求第k大的值,我们就分别用两个数组保留f[v]的前k个值,f[v-c[i]]+w[i]的前k个值,再将这两个数组合并,取第k名. 即f的数组会增加一维. http://blog.csdn.net/lulipeng_cpp/article/details/…

#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> using namespace std; ][],val[],vol[], A[],B[]; int main() { int T,n,v,K,k; scanf("%d",&T); while(T--) { scanf("%d%d%…

The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it doesn't matter,I will give you a link: Here is the link: http://acm.hdu.edu.c…

这题和典型的01背包求最优解不同,是要求第k优解,所以,最直观的想法就是在01背包的基础上再增加一维表示第k大时的价值.具体思路见下面的参考链接,说的很详细 参考连接:http://laiba2004.blog.163.com/blog/static/8835120220138611342496/http://hi.baidu.com/chenyun00/item/1c6c44318acc8bfaa88428c7 #include <iostream> #include <cstdio&…

http://acm.hdu.edu.cn/showproblem.php?pid=2639 题目大意是,往背包里赛骨头,求第K优解,在普通01背包的基础上,增加一维空间,那么F[i,v,k]可以理解为前i个物品,放入容量v的背包时,第K优解的值.时间复杂度为O(NVK). Talk is cheap. 看代码吧. import java.util.Scanner; public class BoneCollector { public static void main(String[] sur…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2639求01背包的第k大解.合并两个有序序列 选取物品i,或不选.最终的结果,是我们能在O(1)的时间内,判定对于体积j,是否应当选取第i件物品. 我们在这里作出了最优的选择.那被我们抛弃的选择呢?他很可能是次优解,第三优解,无论怎样,他都对我们本题求前K优解,起到了重要的作用! #include<stdio.h> #include<string.h> #include<algor…