Lightoj 1088 - Points in Segments 【二分】

题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1088

题意： 有一维的n个点和q条线段。询问每条线段上的点有多少个；

思路：寻找这些点中对于每条线段的上下界即可。

代码：

#include <stdio.h>#include <ctime>#include <math.h>#include <limits.h>#include <complex>#include <string>#include <functional>#include <iterator>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <set>#include <map>#include <list>#include <bitset>#include <sstream>#include <iomanip>#include <fstream>#include <iostream>#include <ctime>#include <cmath>#include <cstring>#include <cstdio>#include <time.h>#include <ctype.h>#include <string.h>#include <assert.h>using namespace std;int n, q;int a[100010];int x, y;int ok;int query(int t,int &ok){    ok = 0;    int left = 1;    int right = n;    int mid;    while (left <= right)    {        int mid = (left + right) / 2;        if (a[mid] >= t)        {            if (a[mid] == t)                ok = 1;            right = mid - 1;        }        else            left = mid + 1;    }    return left;}int main(){    int t;    scanf("%d",&t);    int cases = 1;    while (t--)    {        scanf("%d%d",&n,&q);        for (int i = 1; i <= n; i++)            scanf("%d",&a[i]);        printf("Case %d:\n", cases++);        while (q--)        {            scanf("%d%d", &x, &y);            int ok1 = 0, ok2 = 0, tmp = 0;            int t1 = query(x,ok1);            int t2 = query(y,ok2);            if (ok2) tmp = 1;            //printf("      %d %d\n",t1,t2);            printf("%d\n", t2 - t1 + tmp);        }    }    return 0; }