HDU5178：pairs（二分法 & 尺取法）

pairs

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2497    Accepted Submission(s): 930

Problem Description

John has n points on the X axis, and their coordinates are (x[i],0),(i=0,1,2,…,n−1). He wants to know how many pairs<a,b> that |x[b]−x[a]|≤k.(a<b)

 

Input

The first line contains a single integer T (about 5), indicating the number of cases.
Each test case begins with two integers n,k(1≤n≤100000,1≤k≤109).
Next n lines contain an integer x[i](−109≤x[i]≤109), means the X coordinates.

 

Output

For each case, output an integer means how many pairs<a,b> that |x[b]−x[a]|≤k.

 

Sample Input

2
5 5
-100
0
100
101
102
5 300
-100
0
100
101
102

 

Sample Output

3
10

 

Source

BestCoder Round #31

# include <stdio.h>
# include <algorithm>
using namespace std;
int a[100001];
int main()
{
    int n, t, k;
    scanf("%d",&t);
    while(t--)
    {
        long long sum = 0;
        scanf("%d%d",&n,&k);
        for(int i=0; i<n; ++i)
            scanf("%d",&a[i]);
        sort(a, a+n);
        for(int i=0; i<n; ++i)
        {
            int l = i+1;
            int r = n-1;
            while(l<=r)
            {
                int mid = (l+r)>>1;
                if(a[mid]-a[i] > k)
                    r = mid - 1;
                else
                    l = mid +1;
            }
            sum += r - i;
        }
        printf("%lld\n",sum);
    }
    return 0;
}

使用upper_bound函数版本：

# include <stdio.h>
# include <algorithm>
using namespace std;

int a[100001];
int main()
{
    int n, k, t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        long long ans = 0;
        for(int i=0; i<n; ++i)
            scanf("%d",&a[i]);
        sort(a, a+n);
        for(int i=0; i<n-1; ++i)
        {
            int pos = upper_bound(a+i,a+n, a[i]+k)-a;
            ans += pos - i - 1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}

尺取法版本：

# include <stdio.h>
# include <algorithm>
using namespace std;

int a[100001];
int main()
{
    int n, k, t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        long long ans = 0;
        for(int i=0; i<n; ++i)
            scanf("%d",&a[i]);
        sort(a, a+n);
        for(int l=0,r=0; l<n; ++l)
        {
            while(r+1<n && a[r+1]-a[l]<=k) ++r;
            ans += r-l;
        }
        printf("%lld\n",ans);
    }
    return 0;
}