HDU2639:Bone Collector II(01背包)
2013-04-14 20:54
302 查看
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 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.
Input
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
Sample Output
01背包的第K优解问题
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.
Input
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
01背包的第K优解问题
#include <stdio.h> #include <algorithm> #include <string.h> using namespace std; struct Node { int v,w; } node[1005]; int max(int a,int b) { return a>b?a:b; } int main() { int t; scanf("%d",&t); while(t--) { int n,v,k,j; scanf("%d%d%d",&n,&v,&k); int i; for(i = 0; i<n; i++) scanf("%d",&node[i].w); for(i = 0; i<n; i++) scanf("%d",&node[i].v); int dp[1005][31] = {0}; int a[31],b[31]; for(i = 0; i<n; i++) { for(j = v; j>=node[i].v; j--) { int d; for(d= 1; d<=k; d++) { a[d] = dp[j-node[i].v][d]+node[i].w; b[d] = dp[j][d]; } int x,y,z; x = y = z = 1; a[d] = b[d] = -1; while(z <= k && (y <= k || x <= k)) { if(a[x] > b[y]) dp[j][z] = a[x++]; else dp[j][z] = b[y++]; if(dp[j][z]!=dp[j][z-1]) z++; } } } printf("%d\n",dp[v][k]); } return 0; }
相关文章推荐
- 【01背包-Kth优解】HDU2639-Bone Collector II
- hdu2639 bone collector II 01背包第k优解 TWT Tokyo Olympic 1COMBO-1
- HDU-2639-Bone Collector II(01背包的第k优解)
- hdu 2639 Bone Collector II(01背包)(第k优解)
- HDU 2639 Bone Collector II (01背包,第k解)
- hdu 2639 Bone Collector II(01背包 第K大解)
- HDU2639-Bone Collector II
- hdu2639 Bone Collector II 01背包第K优解
- Bone Collector II----HDU_2639----01背包的变式
- Bone Collector II (HDU_2639) 01背包 + 第K优解
- hdu 2639 Bone Collector II(01背包 第K优解)
- hdu2639 Bone Collector II(背包中第 k 大值)
- Bone Collector II---hdu2639(01背包求第k优解)
- hdu 2639 Bone Collector II (01背包,第k优解问题)
- hdu2639 Bone Collector II--01背包K优解
- hdu2639---Bone Collector II(01背包升华)
- HDU 2639 Bone Collector II (01背包 第K优解)
- hdu Bone Collector II(01背包第K优解)
- hdu 2639 Bone Collector II (01背包)
- HDU2639 Bone Collector II