HDU5124：lines（线段树+离散化）或（离散化思想）
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.
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; }
