http://acm.hdu.edu.cn/showproblem.php?pid=5124
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
2
5
1 2
2 2
2 4
3 4
5 1000
5
11
2 2
3 3
4 4
5 5
Sample Output
3
1
题目解析：
题意：给定 n 个区间，问最多重复的子区间？
题解：（离散化思想）讲所有的数都排个序，将区间的左值定为 1 ，右值定为 -1 ，这样对所有的数搜一遍过去找最大的值即可；或者用线段树+离散化。
一：线段树
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#define N 100010
using namespace std;
struct li
{
int x,y;
}line[N];
struct node
{
int l,r;
int lz;
}q[8*N];
int n,tt,X[2*N];
int maxx;
void build(int l,int r,int rt)
{
q[rt].l=l;
q[rt].r=r;
q[rt].lz=0;
if(l==r)
return ;
int mid=(l+r)>>1;
build(l,mid,rt<<1);
build(mid+1,r,rt<<1|1);
return ;
}
void pushdown(int rt)
{
if(q[rt].lz)
{
q[rt<<1].lz+=q[rt].lz;
q[rt<<1|1].lz+=q[rt].lz;
q[rt].lz=0;
}
}
void update(int lf,int rf,int l,int r,int rt)
{
if(lf<=l&&rf>=r)
{
q[rt].lz+=1;
return ;
}
pushdown(rt);
int mid=(l+r)>>1;
if(lf<=mid) update(lf,rf,l,mid,rt<<1);
if(rf>mid) update(lf,rf,mid+1,r,rt<<1|1);
return ;
}
void query(int l,int r,int rt)
{
if(l==r)
{
maxx=max(maxx,q[rt].lz);
return ;
}
pushdown(rt);
int mid=(l+r)>>1;
query(l,mid,rt<<1);
query(mid+1,r,rt<<1|1);
return ;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
tt=0;
maxx=-1;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d%d",&line[i].x,&line[i].y);
X[tt++]=line[i].x;
X[tt++]=line[i].y;
}
sort(X,X+tt);
int sum=unique(X,X+tt)-X;
build(1,sum,1);
for(int i=0;i<n;i++)
{
int le=lower_bound(X,X+sum,line[i].x)-X+1;
int re=lower_bound(X,X+sum,line[i].y)-X+1;
if(le<=re) update(le,re,1,sum,1);
}
query(1,sum,1);
printf("%d\n",maxx);
}
return 0;
}
二：离散化：思路：可以把一条线段分出两个端点离散化，左端点被标记为-1，右端点被标记为1，然后排序，如果遇到标记为-1，cnt++，否则cnt--;找出cnt的最大值。
#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxn 200010
using namespace std;
struct node
{
int x,c;
bool operator<(const node &a)const
{
return (x<a.x)||(x==a.x&&c<a.c);
}
}p[maxn];
int t;
int n;
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=0; i<n; i++)
{
int s,e;
scanf("%d%d",&s,&e);
p[i*2].x=s;
p[i*2].c=-1;
p[i*2+1].x=e;
p[i*2+1].c=1;
}
sort(p,p+2*n);
int cnt=0; int ans=0;
for(int i=0; i<2*n; i++)
{
if(p[i].c==-1)
{
cnt++;
}
else
cnt--;
ans=max(ans,cnt);
}
printf("%d\n",ans);
}
return 0;
}
可惜做BC时，这两种方法都没想出来，悲催！
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <ctype.h>
#include <string>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <vector>
#include <algorithm>
#include <iostream>
#define PI acos( -1.0 )
using namespace std;
typedef long long ll;
const int NO = 1e5 + 10;
struct ND
{
int x, y;
}st[NO<<1];
int n;
bool cmp( const ND &a, const ND &b )
{
if( a.x == b.x ) return a.y > b.y;
return a.x < b.x;
}
int main()
{
int T;
scanf( "%d",&T );
while( T-- )
{
scanf( "%d", &n );
int cur = 0;
for( int i = 0; i < n; ++i )
{
scanf( "%d", &st[cur].x );
st[cur++].y = 1;
scanf( "%d", &st[cur].x );
st[cur++].y = -1;
}
sort( st, st+cur, cmp );
int ans = 0;
int Max = 0;
for( int i = 0; i < cur; ++i )
{
ans += st[i].y;
Max = max( ans, Max );
}
printf( "%d\n", Max );
}
return 0;
}
