Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5463    Accepted Submission(s): 2880 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4739    Accepted Submission(s): 2470 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took pa…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1334    Accepted Submission(s): 666 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3355    Accepted Submission(s): 1726 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par…

题目链接 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 doesn't matter,I will give you a link: Here is the l…

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…

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/…

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&…

此题就是在01背包问题的基础上求所能获得的第K大的价值. 详细做法是加一维去推当前背包容量第0到K个价值,而这些价值则是由dp[j-w[ i ] ][0到k]和dp[ j ][0到k]得到的,事实上就是2个数组合并之后排序,可是实际做法最好不要怎么做.由于你不知道总共同拥有多少种.而我们最多仅仅须要前K个大的即可了(由于可能2个数组加起来的组合数达不到K个),假设所有加起来数组开多大不清楚,所以能够选用归并排序中把左右2个有序数组合并成一个有序数组的方法来做.就是用2个变量去标记2个有序数组的头…

分析 \(dp[i][j][k]\)为枚举到前i个物品,容量为j的第k大解.则每一次状态转移都要对所有解进行排序选取前第k大的解.用两个数组\(vz1[],vz2[]\)分别记录所有的选择情况,并选择其中前k大的更新当前的dp[i][k].因为dp[i]满足递增的特点,所以可以对两个数组顺序比较选择. #include<bits/stdc++.h> using namespace std; const int maxn = 1e3+5; const int INF = 0x3f3f3f3f;…