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1324: [example 6.6] integer interval

2022-07-07 10:32:00 【A program ape who beats the keyboard violently】

1324：【 example 6.6】 Integer interval

The time limit : 1000 ms Memory limit : 65536 KB
Submission number : 10710 Passing number : 6431

【 Title Description 】

Please program to complete the following tasks ：

1. Read the number of closed intervals and their descriptions ;

2. Find a set with the least number of elements , So that for each interval , At least one integer belongs to the set , Output the number of elements of the collection .

【 Input 】

The first line contains the number of intervals nn,1≤n≤100001≤n≤10000, Next nn That's ok , Each line contains two integers a,ba,b, Separated by a space ,0≤a≤b≤100000≤a≤b≤10000, They are the start and end values of an interval .

【 Output 】

The number of set elements in the first line , For each interval, at least one integer belongs to the set , And the set contains the least number of elements .

【 sample input 】

【 sample output 】

【 Algorithm analysis 】

Algorithm model ： to n Closed interval [ai,bi], Choose as few points as possible on the number axis , Make at least one point in each interval .
Algorithm ： First click b1<=b2<=...<=bn Sort . Mark the right end of the current interval every time n, And move the current interval pointer to the right , Until the current interval does not contain n, Repeat the above operation .

【AC Code 】

#include<algorithm>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<string>
#include<vector>
using namespace std;
const int N=1e4+10;
inline int fread()
{
	char ch=getchar();
	int n=0,m=1;
	while(ch<'0' or ch>'9')
	{
		if(ch=='-')m=-1;
		ch=getchar();
	}
	while(ch>='0' and ch<='9')n=(n<<3)+(n<<1)+ch-48,ch=getchar();
	return n*m;
}
void fwrite(int n)
{
	if(n>9)fwrite(n/10);
	putchar(n%10+'0');
}
int n,a[N],b[N],ans,m;
void qsort(int n,int m)// Multi keyword quick sorting 
{
	int i=n,j=m,mid=b[n+m>>1],mid1=a[n+m>>1];
	while(i<=j)
	{
		while(b[i]<mid or(b[i]==mid and a[i]<mid1))i++;
		while(b[j]>mid or(b[j]==mid and a[j]>mid1))j--;
		if(i<=j)swap(b[i],b[j]),swap(a[i],a[j]),i++,j--;
	}
	if(n<j)qsort(n,j);
	if(i<m)qsort(i,m);
}
signed main()
{
	n=fread();
	for(int i=1;i<=n;i++)a[i]=fread(),b[i]=fread();
	qsort(1,n);
	for(int i=1,j=-1;i<=n;i++)// While initializing the loop variable , initialization j
	{// Make j=-1 You can make the operation of the first interval the same as that of other intervals  
		if(j>=a[i])continue;// If the current interval contains marked points , Just skip. 
		ans++,j=b[i];// Update marked points 
	}
	fwrite(ans);
	return 0;
}