HDU2639Bone Collector II[01背包第k优值]
2016-09-22 23:06
190 查看
Bone Collector II
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4229 Accepted Submission(s): 2205
[align=left]Problem Description[/align]
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.cn/showproblem.php?pid=2602
Today we are not desiring the maximum value of bones,but the K-th maximum value of the bones.NOTICE that,we considerate two ways that get the same value of bones are the same.That means,it will be a strictly decreasing sequence from the 1st maximum , 2nd maximum .. to the K-th maximum.
If the total number of different values is less than K,just ouput 0.
[align=left]Input[/align]
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, K(N <= 100 , V <= 1000 , K <= 30)representing the number of bones and the volume of his bag and the K we need. 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 K-th maximum of the total value (this number will be less than 231).
Sample Input
3
5 10 2
1 2 3 4 5
5 4 3 2 1
5 10 12
1 2 3 4 5
5 4 3 2 1
5 10 16
1 2 3 4 5
5 4 3 2 1
Sample Output
12
2
0
Author
teddy
每个状态保存1到k优值,转移时给f[j][]和f[j-v][]+w二路归并一下就好了
复杂度O(nvk)
注意本题是严格递减
// // main.cpp // hdu2639 // // Created by Candy on 9/22/16. // Copyright © 2016 Candy. All rights reserved. // #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; const int N=105,V=1005,K=35,INF=1e9+5; int read(){ char c=getchar();int x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();} return x*f; } int T,n,m,k,w ,v ; int f[V][K],a[K],b[K]; void dp(){ memset(f,0,sizeof(f)); memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); for(int i=1;i<=n;i++) for(int j=m;j>=v[i];j--){ for(int z=1;z<=k;z++) {a[z]=f[j][z];b[z]=f[j-v[i]][z]+w[i];} int x=1,y=1,z=1; while(z<=k&&(x<=k||y<=k)){ if(a[x]>=b[y]) f[j][z]=a[x++]; else f[j][z]=b[y++]; if(f[j][z]!=f[j][z-1]) z++; } } } int main(int argc, const char * argv[]) { T=read(); while(T--){ n=read();m=read();k=read(); for(int i=1;i<=n;i++) w[i]=read(); for(int i=1;i<=n;i++) v[i]=read(); dp(); printf("%d\n",f[m][k]); } return 0; }
相关文章推荐
- 01背包(第k优值)
- dp之01背包hdu2639(第k优解)
- hdu 2636 Bone Collector II 01背包(第k优解问题)
- 01背包 第k优解
- 2639骨头问题2(01背包的第k大值)
- hdu2639 bone collector II 01背包第k优解 TWT Tokyo Olympic 1COMBO-1
- HDOJ2639(01背包第k最优解模板题)
- hdu2639-01背包(第k大背包问题)
- hdu 2639 (第k小的01背包)
- hdu2639——Bone Collector II——————【01背包、第k优解】
- HDU 2639 Bone Collector II (01背包,第k解)
- Bone Collector II HDU - 2639 01背包,第k优解
- hdu(2639)(01背包的第k最优解)
- hdu2639(第k大值+01背包)
- hdu 2639 Bone Collector II(01背包)(第k优解)
- hdu 2639 Bone Collector II (01背包,第k优解问题)
- HDU 2639 Bone Collector II 【01背包的第k优解】
- HDU 2639(01背包第K大)
- hdu2639(01背包的第K大)
- hdu 2639 Bone Collector II(01背包的第k值)