HDOJ 4016 Magic Bitwise And Operation(Dfs剪枝）

if we do no cutting edge strategy in this dfs problem,we will get a TLE,so here is the cutting edge : if we find a solution that is bigger than ans we have after doing the AND operation to the end , we can just cut down from the current number. So we need to get a result of suffix AND sum from all the number we get.

#include <bits/stdc++.h>
#include <cstring>
#define ll long long
#define inf 0x7fffffffffffffff
using namespace std;
ll num[50],a[50];
ll ans;
int n,k;
int vis[50];
void dfs(int pos,ll pre,int step)
{
// cout<<pos<<" "<<pre<<" "<<step<<endl;
if(pre < ans)
{
ans = pre;
}
if((pre & a[pos]) >= ans)
{
return;
}
if(pos == n || step == k)
{
return ;
}
dfs(pos+1,pre,step);
dfs(pos+1,pre&num[pos],step+1);

}
int main()
{
int ca,cat = 1;
cin>>ca;
while(ca--)
{
cin>>n>>k;
for(int i=0;i<n;i++)
{
cin>>num[i];
}
sort(num,num+n);
a[n-1] = num[n-1];
for(int i=n-2;i>= 0;i--)
{
a[i] = num[i] & a[i+1];
}
memset(vis,0,sizeof(vis));
ans = inf;
dfs(0,inf,0);
cout<<"Case #"<<cat++<<": ";
cout<<ans<<endl;
}
return 0;
}