思路：考虑一个任务的“最晚开始时间”即任务的起止时间的中点（ startTime + (endTime-startTime)/2 ），因为如果该任务过了最晚开始时间仍然没有开始执行，那么肯定无法完成该任务，所以按照最晚开始时间进行排序。

代码一：

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

struct Time{
	Time(){}
	Time(int s,int e,int m){
		startTime = s;
		endTime = e;
		midTime = m;
	}
	int startTime;
	int endTime;
	int midTime;
};

bool cmp(Time a, Time b){
	return a.midTime < b.midTime;
}

int main(){	
	int n,i,now;
	cin>>n;
	vector<Time> vec(n);
	int start,end,mid;
	for(i=0; i<n; i++){	
		cin>>start>>end; 
		mid = start + (end - start)/2;	 	 
		vec[i].startTime = start;
		vec[i].endTime = end;
		vec[i].midTime = mid;
	}	
	sort(vec.begin(), vec.end(), cmp);
	bool attend = true;
	now = 0;
	for(i=0; i<vec.size(); i++){
		if( now < vec[i].startTime ){
		      now = vec[i].startTime;
		}
		now += (vec[i].endTime-vec[i].startTime)/2 + 1 ;
		if(now > vec[i].endTime){
			attend = false;
			break;
		}
	}	
	if(attend){
		cout<<"YES"<<endl;
	}else{
		cout<<"NO"<<endl;
	}
	return 0;
}

#include <iostream>
#include<algorithm>
using namespace std;
struct point {
	int s, t;
} p[10100];
int now, i, n;
bool boo;
bool cmp(point a, point b) {
	return (a.t + a.s) < (b.t + b.s);
}
int main() {
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	scanf("%d", &n);
	for (i = 0; i < n; i++) {
		scanf("%d%d", &p[i].s, &p[i].t);
	}
	boo = true;
	sort(p, p + n, cmp);
	now = 0;
	for (i = 0; i < n; i++) {
		if (now < p[i].s)
			now = p[i].s;
		now += (p[i].t - p[i].s) / 2 + 1;
		if (now > p[i].t)
			boo = false;
	}
	if (boo)
		printf("YES\n");
	else
		printf("NO\n");
	return 0;
}

        

任务排序