51nod

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1278

因为圆心都在x轴上,把每个圆转化成线段后,按线段的起点排序,那么对于每个圆都要从后面找出起点大于当前圆转化成的线段终点的一个点,这个点之后的圆都会与当前圆相离.

因为按起点排序所以可以二分求解.

发现自己二分写的乱七八糟的.

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>
#pragma comment(linker, "/STACK:102400000,102400000")
#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define ll long long
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)

#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d
", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("a.txt", "r", stdin)
#define Write() freopen("b.txt", "w", stdout);
#define maxn 1000000000
#define N 2510
#define mod 1000000000
using namespace std;

struct node
{
    int x,y;
    bool operator < (const node a) const
    {
        if(x==a.x) return y<a.y;
        return x<a.x;
    }
}p[50001];
int n;
int erfen(int a,int b)
{
    int l=a,r=n-1,mid=0;
    while(l<=r)
    {
        mid=(l+r)/2;
        if(p[mid].x>b) r=mid-1;
        else l=mid+1;
    }
    if(p[mid].x<=b) mid++;
    if(mid>=n) return n;
    else return mid;
}

int main()
{
    //freopen("a.txt","r",stdin);
    int a,b;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&a,&b);
        p[i].x=a-b;
        p[i].y=a+b;
    }
    sort(p,p+n);
    //for(int i=0;i<n;i++)
    //{
      //  printf("%d %d
",p[i].x,p[i].y);
    //}
    int num=0;
    for(int i=0;i<n-1;i++)
    {
        int x=erfen(i+1,p[i].y);
        num+=n-x;
       // printf("%d %d
",num,x);

    }
    printf("%d
",num);
    return 0;
}

当然也可以直接把p[i].x赋值给另外一个一维数组,然后就可以用upper_bound实现二分查找了.