Bone Collector II Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4542 Accepted Submission(s): 2365
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 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
题目和前面收集石子那个有一点出处，本题是求第k大的价值是多少 输入n个val和weight，求背包为m大时，第k大价值是多少 dp[i][j]表示前i个,背包空间为j时的最大价值 那么就不用01背包的转移方程dp[i][j]=dp[i][j,dp[i-weight[i]][j]+val[i];了 用两个数组装上两个值，然后再for所有的情况，求出所有的并排序，输入dp[m][k]即可
#include <cstdio> #include <algorithm> #include <cstring> #include <iostream> using namespace std; int main() { int t; int v,m,k,p; int x,y,z; int val[110],weight[110]; int d1[1100],d2[1100]; int dp[1100][110]; while(scanf("%d",&t)==1) { while(t--) { memset(dp,0,sizeof(dp)); memset(val,0,sizeof(val)); memset(weight,0,sizeof(weight)); memset(d1,0,sizeof(d1)); memset(d2,0,sizeof(d2)); scanf("%d%d%d",&v,&m,&k); for(int i=1; i<=v; i++) scanf("%d",&val[i]); for(int i=1; i<=v; i++) scanf("%d",&weight[i]); for(int i=1; i<=v; i++) { for(int j=m; j>=weight[i]; j--) { for(p=1; p<=k; p++) { d1[p]=dp[j][p]; d2[p]=dp[j-weight[i]][p]+val[i]; } d1[p]=d2[p]=-1; x=y=z=1; while((d1[x]!=-1||d2[y]!=-1)&&z<=k) { if(d1[x]>d2[y]) { dp[j][z]=d1[x]; x++; } else { dp[j][z]=d2[y]; y++; } if(dp[j][z-1]!=dp[j][z]) z++; } } } cout<<dp[m][k]<<endl; } } return 0; }