Bone Collector II

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4658    Accepted Submission(s): 2425 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 2 ³¹).   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   Source 百万秦关终属楚   Recommend teddy   |   We have carefully selected several similar problems for you:   2159  2955  1114  3033  1561    典型的0-1背包求k优解问题，就是在每一个状态下面加一个有序队列，把他这种状态所有可能的结果放进去，然后最后输出最后一个状态的第k个解就好了。

我这种菜鸟就不在这里卖弄了，推荐个别人的博客：  点击打开链接   （ http://blog.csdn.net/lulipeng_cpp/article/details/7584981）

ac代码：

#include <stdio.h> #include <math.h> #include <string.h> #include <algorithm> using namespace std; int _max(int x,int y){ return x>y?x:y; } int main(){ int t,n,v,k; int val[105],vol[105]; int dp[1005][35],a[1005],b[1005]; int cnt1,cnt2,cnt3; scanf("%d",&t); while(t--){ scanf("%d%d%d",&n,&v,&k); for(int i=1;i<=n;i++){ scanf("%d",val+i); } for(int i=1;i<=n;i++){ scanf("%d",vol+i); } memset(dp,0,sizeof(dp)); memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); for( int i=1;i<=n;i++ ){ for(int j=v;j>=vol[i];j--){ int kk; for(kk=1;kk<=k;kk++){ a[kk]=dp[j][kk]; b[kk]=dp[j-vol[i]][kk]+val[i]; } a[kk]=-1; b[kk]=-1; cnt1=cnt2=cnt3=1; while( cnt3<=k && (a[cnt1]!=-1 || b[cnt2]!=-1) ){ if(a[cnt1] > b[cnt2]){ dp[j][cnt3]=a[cnt1++]; } else{ dp[j][cnt3]=b[cnt2++]; } if(dp[j][cnt3]!=dp[j][cnt3-1]){ cnt3++; } } // dp[j]=_max(dp[j],dp[j-vol[i]+val[i]); } } printf("%d\n",dp[v][k]); } return 0; }