题目

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e6+10;
int t;
int n; 
struct node{
	int x,y,tx,ty;
}a[N];
bool cmp(node p1,node p2)
{
	if(p1.x==p2.x)
	{
		return p1.y<p2.y;
	}
	return p1.x<p2.x;
}
bool cp(node p1,node p2)
{
	if(p1.y==p2.y)
	{
		return p1.x<p2.x;
	}
	return p1.y<p2.y;
}
int mix,miy;
int nowx,nowy;
queue<node> q;
signed main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin>>t;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
    	cin>>a[i].x>>a[i].y>>a[i].tx>>a[i].ty;
	}
	sort(a+1,a+1+n,cmp);
	mix=a[1].x;
	nowx=mix+a[1].tx;
	for(int i=2;i<=n;i++)
	{
		if(nowx<a[i].x)
		{
			mix+=a[i].x-nowx;
			nowx=a[i].x;
		}
		nowx+=a[i].tx;
	}
	sort(a+1,a+1+n,cp);
	nowx=mix;
	nowy=0;
	miy=0;
	for(int i=1;i<=n;i++)
	{
		q.push(a[i]);
	}
	while(!q.empty())
	{
		node p=q.front();
		q.pop();
		if(nowx<p.x)
		{
			q.push(p);
			continue;
		}
		if(nowy<p.y)
		{
			miy+=p.y-nowy;
			nowy=p.y;
		}
		nowx+=p.tx;
		nowy+=p.ty;
	}
	cout<<mix<<" "<<miy;
  	return 0;
}

思路是先保证力量最小，然后依靠力量最小去找精神最小，求调。