Bone Collector II(HDU 2639 DP)
Bone Collector II
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3471 Accepted Submission(s): 1792
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.
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.
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
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
int dp[][];
int val[],vol[];
int n,v,K;
int a[],b[];
int main()
{
freopen("in.txt","r",stdin);
int i,j,k;
int x,y,z;
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&v,&K);
for(i=;i<=n;i++)
scanf("%d",&val[i]);
for(i=;i<=n;i++)
scanf("%d",&vol[i]);
memset(dp,,sizeof(dp));
for(i=;i<=n;i++)
{
for(j=v;j>=vol[i];j--)
{
for(k=;k<=K;k++) //记录每种状态的k优解
{
a[k]=dp[j][k];
b[k]=dp[j-vol[i]][k]+val[i];
}
a[k]=b[k]=-; //存储的内容已经按照从小到大排序好了,然后合并到dp数组中去
x=y=z=;
while(z<=K&&(a[x]!=-||b[y]!=-))
{
if(a[x]>b[y])
{
dp[j][z]=a[x];
x++;
}
else
{
dp[j][z]=b[y];
y++;
}
if(dp[j][z-]!=dp[j][z])
z++;
}
}
}
printf("%d\n",dp[v][K]);
}
return ;
}
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