01背包-第k优解
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
#include <iostream>
#include <cstdio>
using namespace std;
#define max(a,b) ((a)>(b)?(a):(b))
const int maxn = ;
int main()
{
int T;
scanf("%d", &T);
int dp[maxn][], val[maxn], vol[maxn], A[], B[];
while (T--)
{
int n, v, k;
scanf("%d %d %d", &n, &v, &k);
int i, j, kk;
for (i=; i<n; i++) scanf("%d", &val[i]);
for (i=; i<n; i++) scanf("%d", &vol[i]);
memset(dp, , sizeof(dp)); int a, b, c;
for (i=; i<n; i++)
for (j=v; j>=vol[i]; j--)
{
for (kk=; kk<=k; kk++)
{
A[kk] = dp[j-vol[i]][kk] + val[i];
B[kk] = dp[j][kk];
}
A[kk] = -, B[kk] = -;
a = b = c = ;
while (c<=k && (A[a] != - || B[b] != -))
{
if (A[a] > B[b])
dp[j][c] = A[a++];
else
dp[j][c] = B[b++];
if (dp[j][c] != dp[j][c-])
c++;
}
} printf("%d\n", dp[v][k]);
}
return ;
}
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