【CUGBACM15级BC第20场 B】hdu 5124 lines

lines

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1837    Accepted Submission(s): 782

Problem Description

John has several lines. The lines are covered on the X axis. Let A is a point which is covered by the most lines. John wants to know how many lines cover A.

 

Input

The first line contains a single integerT(1≤T≤100)(the data for N>100 less than 11 cases),indicating the number of test cases.
Each test case begins with an integer N(1≤N≤105),indicating the number of lines.
Next N lines contains two integers Xi and Yi(1≤Xi≤Yi≤109),describing a line.

 

Output

For each case, output an integer means how many lines cover A.

 

Sample Input

251 2 2 22 43 45 100051 12 23 34 45 5

 

Sample Output

31

 

题意：x轴上很多段区间，求覆盖最多的一个点的覆盖次数

思路：首先很容易想到这个点一定可以是某一段区间的端点，然后这里给了区间，要求一个点的出现次数，想到什么？

没错，区间更新，单点求和。 当然用线段树更加直观，只是本人确实不是很喜欢线段树...于是就用树状数组实现了

与单点求和，区间求和的区别在于更新区间的时候add(l,1)和add(r+1,-1)  单点求和就是直接sum(i)就可以了

记得要离散化处理

#include <iostream>#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <algorithm>#include <functional>#define PI acos(-1)#define eps 1e-8#define inf 0x3f3f3f3f#define debug(x) cout<<"---"<<x<<"---"<<endltypedef long long ll;using namespace std;const int N = 1e5 + 10;int a[N], b[N], p[2 * N], maxn;int lowbit(int x){    return x & (-x);}///区间修改+单点查询int arr[N * 2];inline int sum(int x){    int res = 0;    while (x)    {        res += arr[x], x -= lowbit(x);    }    return res;}inline void add(int x, int n){    while (x < maxn)    {        arr[x] += n, x += lowbit(x);    }}inline void update(int x, int y, int n)///区间修改{    add(x, n);    add(y + 1, -n);}int main(){    std::ios::sync_with_stdio(false);    int t;    cin >> t;    while (t--)    {        int n;        cin >> n;        memset(arr, 0, sizeof(arr));        maxn = 2 * n;        for (int i = 1; i <= n; i++)        {            cin >> a[i] >> b[i];            p[i] = a[i];            p[i + n] = b[i];        }        sort(p + 1, p + 1 + 2 * n);        for (int i = 1; i <= n; i++)        {            int l = lower_bound(p + 1, p + 1 + 2 * n, a[i]) - p;            int r = lower_bound(p + 1, p + 1 + 2 * n, b[i]) - p;            update(l, r, 1);        }        int ans = 1;        for (int i = 1; i <= 2 * n; i++)        {            int s = sum(i);            ans = max(ans, s);        }        cout << ans << endl;    }    return 0;}