Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3437    Accepted Submission(s): 1773 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par…

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…

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…

http://acm.hdu.edu.cn/showproblem.php?pid=2639 http://blog.csdn.net/lulipeng_cpp/article/details/7584981 求第K大的思路是把每个d[v]看成是由d[v]和d[v-cost]+weight两个序列组成的,然后分别记录每个序列的第k大,然后逐项更新. 用个形象的比喻吧:如果我想知道学年最高分,那么,我只要知道每个班级的最高分,然后统计一遍就可以了. 如果我想知道学年前十呢?我必须要知道每个班的前十…

#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%…

/* 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…

题目链接[http://acm.hdu.edu.cn/showproblem.php?pid=2639] 题意:求第k大背包. 题解:利用二路归并的思想,求解第K大的值. #include<bits/stdc++.h> using namespace std; typedef long long LL; ; ], w[MAXN], v[MAXN]; int N, W, K; ], B[]; void super_kth() { ; i <= N; i++) { for(int j = W…

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/…

题意: 01背包,找出第k最优解 题解: 对于01背包最优解我们肯定都很熟悉 第k最优解的话也就是在dp方程上加一个维度来存它的第k最优解(dp[i][j]代表,体积为i能获得的第j最大价值) 对于每一个物品只有两种选择情况 1.把这个物品加入背包 2.不要这个物品 那么它的前k种最优解也是由n种物品的这两个选择组成的 假设总体积是4,现在有两种物品,求第2最大值 1.体积1,价值3 2.体积1,价值2 最开始dp状态: 体积:   1  2  3  4 第1最大值:0  0  0  0 第2最…

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…