题目描述
There are N islands lining up from west to east, connected by N−1 bridges.
The i-th bridge connects the i-th island from the west and the (i+1)-th island from the west.
One day, disputes took place between some islands, and there were M requests from the inhabitants of the islands:
Request i: A dispute took place between the ai-th island from the west and the bi-th island from the west. Please make traveling between these islands with bridges impossible.
You decided to remove some bridges to meet all these M requests.
Find the minimum number of bridges that must be removed.
Constraints
·All values in input are integers.
·2≤N≤105
·1≤M≤105
·1≤ai<bi≤N
·All pairs (ai,bi) are distinct.
输入
Input is given from Standard Input in the following format:
N M
a1 b1
a2 b2
:
aM bM
输出
Print the minimum number of bridges that must be removed.
样例输入
5 2
1 4
2 5
样例输出
1
提示
The requests can be met by removing the bridge connecting the second and third islands from the west.
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
#define end return 0
struct node
{
int l;
int r;
};
node a[100500]= {0};
bool cmp(node a,node b)
{
if(a.l!=b.l)
return a.l<b.l;
else
return a.r<b.r;
}
ll sum=1;
int main()
{
int n=read,m=read;
for(int i=0;i<m;i++)
a[i].l=read,a[i].r=read;
sort(a,a+m,cmp);
int minn=a[0].r;
for(int i=1;i<m;i++)
{
if(a[i].l>=minn)
{
sum++;
minn=a[i].r;
}
else
minn=min(a[i].r,minn);
}
printf("%lld
",sum);
return 0;
}
/**************************************************************
Language: C++
Result: 正确
Time:56 ms
Memory:2812 kb
****************************************************************/