Java(javac 1.8) 解法, 执行用时: 1171ms, 内存消耗: 46040K, 提交时间: 2020-05-30 13:01:25

import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		try (Scanner sc = new Scanner(System.in)) {
			int N[][] = new int[sc.nextInt()][], max[] = new int[N.length], min[] = new int[N.length], count = 0;
			for (int i = 0; i < N.length; i++) {
				N[i] = new int[] { sc.nextInt(), sc.nextInt() };
			}
			Arrays.sort(N, (a, b) -> a[0] - b[0]);
			for (int i = 0; i < N.length; i++) {
				max[i] = i > 0 ? Math.max(max[i - 1], N[i][1]) : N[i][1];
			}
			for (int i = N.length - 1; i >= 0; i--) {
				min[i] = i < N.length - 1 ? Math.min(min[i + 1], N[i][1]) : N[i][1];
			}
			for (int i = 0; i < N.length; i++) {
				if ((i == 0 || max[i - 1] < N[i][1]) && (i == N.length - 1 || min[i + 1] > N[i][1])) {
					count++;
				}
			}
			System.out.println(count);
		}
	}
}

C++14(g++5.4) 解法, 执行用时: 49ms, 内存消耗: 2028K, 提交时间: 2020-07-05 15:27:48

#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
const int oo = 0x3f3f3f3f;
int n, mxpre[N], mnsuf[N], ans, i;
struct Rana
{
    int a, b;
} p[N];
int main()
{
    scanf("%d", &n);
    for (i = 1; i <= n; ++i)
        scanf("%d%d", &p[i].a, &p[i].b);
    sort(p + 1, p + n + 1, [](const Rana u, const Rana v) { return u.a < v.a; });
    mxpre[0] = -oo;
    for (i = 1; i <= n; ++i)
        mxpre[i] = max(mxpre[i - 1], p[i].b);
    mnsuf[n + 1] = oo;
    for (i = n; i > 0; --i)
        mnsuf[i] = min(mnsuf[i + 1], p[i].b);
    for (i = 1; i <= n; ++i)
        if (mxpre[i - 1] < p[i].b && p[i].b < mnsuf[i + 1])
            ++ans;
    printf("%d", ans);
}

C++11(clang++ 3.9) 解法, 执行用时: 53ms, 内存消耗: 1260K, 提交时间: 2020-05-30 09:47:37

#include<algorithm>
#include<cstdio>
using namespace std;
struct node{int x,y;}a[100010];
bool cmp(node a,node b){return a.x<b.x;}
int maxx=-1000000000,ans=0,n;bool h[100010];
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i)scanf("%d%d",&a[i].x,&a[i].y);
	sort(a+1,a+n+1,cmp);
	for(int i=1;i<=n;++i)if(maxx>a[i].y)h[i]=1;else maxx=a[i].y;
	for(int i=n;i>=1;--i)if(maxx<a[i].y)h[i]=1;else maxx=a[i].y;
	for(int i=1;i<=n;++i)if(!h[i])ans++;
	printf("%d\n",ans);
 	return 0;
}