HDU 2639 Bone Collector II

Problem DescriptionThe Title of this problem be familiar,isn ' t it?yeah,if you had took part in the ' Rookie Cup ' competitio N,you must has seem this title. If you haven ' t seen it before,it doesn ' t matter,i would give you a link:

The link:http://acm.hdu.edu.cn/showproblem.php?pid=2602

Today We is not desiring the maximum value of bones,but the k-th maximum value of the bones. NOTICE That,we considerate and ways that get the same value of bones is the same. That means,it'll be a strictly decreasing sequence from the 1st maximum, 2nd maximum. To the k-th maximum.

If the total number of different values are less than k,just ouput 0.
Inputthe first line contain a integer T, the number of cases.
Followed by T cases, each case three lines, the first line contain both integer n, v, K (n <=, v <=, K & lt;=) 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 for each bone.

Outputone integer per line representing the k-th maximum of the total value (this number would be less than 231).

Sample Input

35 10 21 2 3 4 55 4 3 2 15 10 121 2 3 4 55 4 3 2 15 10 161 2 3 4 55 4 3 2 1

Sample Output

122
0
For the Big K. Start thinking is to add a big to small to find the value of K, why not
Add one-dimensional value that represents the K-large when the volume is J. Use two arrays to store the two states in addition or not, and merge with the idea of merging.
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>using namespace    STD; #define INF 0x3f3f3f3fint n,v,k;int va[105],vo[105],dp[1010][50];int a[50],b[50];int main () {int t,i,j,x,y,z,r;    scanf ("%d", &t);        while (t--) {scanf ("%d%d%d", &n,&v,&k);        for (i=0;i<n;i++) scanf ("%d", &va[i]);        for (i=0;i<n;i++) scanf ("%d", &vo[i]);        memset (dp,0,sizeof DP);                    for (i=0;i<n;i++) for (j=v;j>=vo[i];j--) {for (r=1;r<=k;r++) {                    A[r]=dp[j-vo[i]][r]+va[i];                B[R]=DP[J][R];                } a[r]=-1,b[r]=-1;                X=y=z=1; while (z<=k&& (a[x]!=-1| |                    B[y]!=-1) {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;}

HDU 2639 Bone Collector II