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New~ 关于举办杭电程序设计竞赛（2014'12）的报名通知 
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lines

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 382    Accepted Submission(s): 187

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 integer T(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

 

Source

BestCoder Round #20

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=100010;int X[maxn*2],N;struct node{    int x,y;    node(){}    node(int l,int r):x(l),y(r){}}line[maxn];struct IntervalTree{    int sum[maxn<<3],add[maxn<<3];    void build(int o,int l,int r)    {        sum[o]=add[o]=0;        if(l==r)return;        int mid=(l+r)>>1;        build(o<<1,l,mid);        build(o<<1|1,mid+1,r);    }    void pushdown(int o)    {        if(add[o])        {            add[o<<1]+=add[o];            add[o<<1|1]+=add[o];            sum[o<<1]+=add[o];            sum[o<<1|1]+=add[o];            add[o]=0;        }    }    void pushup(int o)    {        sum[o]=max(sum[o<<1],sum[o<<1|1]);    }    void update(int o,int l,int r,int q1,int q2,int val)    {        if(q1<=l&&r<=q2)        {            add[o]+=val;            sum[o]+=val;            return ;        }        pushdown(o);        int mid=(l+r)>>1;        if(q1<=mid)update(o<<1,l,mid,q1,q2,val);        if(q2>mid)update(o<<1|1,mid+1,r,q1,q2,val);        pushup(o);    }    int query(int o,int l,int r,int q1,int q2)    {        if(q1<=l&&r<=q2)return sum[o];        pushdown(o);        int mid=(l+r)>>1;        int ans=0;        if(q1<=mid)ans=max(ans,query(o<<1,l,mid,q1,q2));        if(q2>mid) ans=max(ans,query(o<<1|1,mid+1,r,q1,q2));        return ans;    }}tree;int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&N);        int cnt=0;        for(int i=1;i<=N;i++)        {            scanf("%d%d",&line[i].x,&line[i].y);            X[cnt++]=line[i].x,X[cnt++]=line[i].y;        }        sort(X,X+cnt);        cnt=unique(X,X+cnt)-X;        tree.build(1,1,cnt);        for(int i=1;i<=N;i++)        {            int l=lower_bound(X,X+cnt,line[i].x)-X+1;            int r=lower_bound(X,X+cnt,line[i].y)-X+1;            if(l<=r)tree.update(1,1,cnt,l,r,1);        }        printf("%d\n",tree.query(1,1,cnt,1,cnt));    }    return 0;}