来源:POJ3579
两次二分
代码:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int MAXN = 100000+10;
int n;
int a[MAXN];
bool judge(int m){
int cnt=0;
for(int i=0;i<n;++i){
cnt+=n-(lower_bound(a,a+n,a[i]+m)-a);
}
//cout<<"infun: "<<cnt<<endl;
if(cnt>(n-1)*n/4) return 1;
else return 0;
}
int main(){
while(scanf("%d",&n)!=EOF){
for(int i=0;i<n;i++){
scanf("%d",a+i);
}
//cout<<"des: "<<n*(n-1)/4<<endl;
sort(a,a+n);
int left=0;
int right=a[n-1]-a[0];
int mid,ans;
while(left<=right){
mid=(left+right)/2;
//cout<<"pro: "<<left<<" "<<mid<<" "<<right<<endl;
if(judge(mid)){
left = mid+1;
ans=mid;
}
else{
right=mid-1;
}
}
//cout<<left<<endl;
//cout<<mid<<endl;
cout<<right<<endl;
//cout<<ans<<endl;
}
return 0;
}
两次二分解决配对问题

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