【01背包求第K优解】HDU 2639 Bone Collector II

题意：01背包求第k优解

题解：因为01背包其实把所有可能的情况全列举了，取第k优解无非将数组增加一维，f[i][v][k]表示在V体积里前i个物品第k大的解。在求解过程中再每次运算时，运算k次即可，过程需要两个额外数组维护。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <map>
#include <queue>
#include <algorithm>
using namespace std;
const int N = 1010;
const int INF = 0x3f3f3f3f;

int f
[35];
int v[105],w[105],A[35],B[35];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,W,k;
scanf("%d%d%d",&n,&W,&k);
memset(f,0,sizeof(f));
for(int i = 1; i <= n; i++)
scanf("%d",&v[i]);
for(int i = 1; i <= n; i++)
scanf("%d",&w[i]);
for(int i = 1; i <= n; i++)
{
for(int j = W; j >= w[i]; j--)
{
int p;
for(p = 1; p <= k; p++)
{
A[p] = f[j][p];
B[p] = f[j-w[i]][p]+v[i];
}
A[p] = B[p] = -1;
int a,b,c;
a = b = c = 1;
while(c <= k && (A[a] != -1 || B[b] != -1))
{
if(A[a] > B[b])
f[j][c] = A[a++];
else
f[j][c] = B[b++];
if(f[j][c] != f[j][c-1])
c++;
}
}
}
printf("%d\n",f[W][k]);
}
return 0;
}