CodeForces - 589D —（思维题）

Welcoming autumn evening is the best for walking along the boulevard and npeople decided to do so.

The boulevard can be represented as the axis Ox. For every person there are three parameters characterizing the behavior: t_i, s_i, f_i — the moment of time when the i-th person starts walking, the start point and the end point of the walk respectively. Each person moves in a straight line along the boulevard from s_i to f_i with a constant speed of either 1 or - 1 depending on the direction.

When the i-th person appears on the boulevard at the point s_i she immediately starts walking towards the point f_i.

If two or more persons meet at the boulevard (they are at the same point at the same time, no matter which directions they are going) they all greet each other. Like in the normal life, every pair of people greet each other at most once.

You task is to calculate for every person how many people she greets while walking along the boulevard.

Please, pay attention to the fact that i-th person may meet and greet any other person at points s_i and f_i. After a person achieves the destination point f_ishe moves out of the boulevard and cannot greet anyone else. The same rule applies to the start of the walk: a person cannot greet anyone until she appears on the boulevard.

Input

In the first line there is an integer n (2 ≤ n ≤ 1000) — the number of people who decided to go for a walk.

The following n lines contain parameters for n people. In the i-th line there are three positive integers t_i, s_i, f_i (1 ≤ t_i, s_i, f_i ≤ 10⁶, s_i ≠ f_i), where t_i, s_i, f_i — the moment of time when the i-th person starts walking, the start point and the end point of the walk respectively.

Output

The single line of the output should contain a sequence of n integers r₁, r₂, ..., r_n separated by a space, where r_i denotes the number which the i-th person greets other people while walking along the boulevard.

Examples

Input

Output

2 1 1

Input

Output

2 2 2

题意：一个数字n表示有n个人要走上一条路，这条路可以看成一条数轴，每个人用3个数据t，s，f描述，t表示上路的时间，s表示上路的起点，f表示走到的终点（起点可能大于终点，表示往左走），问这n个人每个人会与多少人相遇（走到同一点或互相穿过算相遇）；

思路：每个人的出发时间都不相同，不方便计算，那我们可以预处理一下，将所有点转化为出发时间相同。如果一个人是往左走（起点大于终点），那就将他的出发地点加上时间，向右走就将起点减去时间。这样，就可以将所有的点都看成是从0时刻出发的了。
预处理完之后，用一个for循环的嵌套进行两两遍历，没两个点之间就分两种情况：
1.两个点往相同的方向走：如果要能够相遇，就表示他们预处理完后的起点要相等，因为预处理完后，所有点出发时间都是0，如果起点不同，那两个点同时向一个方向移动，将永远不会相遇。、
而起点相同之后，还要判断两个点的行走区间是否有交集（这里的区间是指未初始化前的区间），有交集才表示会一同走到某一点后相遇。

2.两点的方向相反：若两点的方向相反，我们就可以求出他们相遇的点，就是两点预处理后的起点相加除以2，若求出的相遇的点在两个点行走的区间内，就表示能相遇。
这里的相遇还有一点比较特殊，那就是他们并不一定会走到同一点，比如两个点，一个在2位置，一个在3位置，下一秒，他们就互换了位置，而不会出现在同一点，这也算相遇。如果用整数（2+3）/2，将会得到2，结果可能会出错，所以我们可以用double型，得到一个小数，如果这个小数在两点的交集内，则是可行的。

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
#define eps 1e-7
#define ll long long
#define inf 0x3f3f3f3f
#define pi 3.141592653589793238462643383279
using namespace std;
int main()
{
    double a[1007],b[1007],c[1007];
    int n,visit[1007];
    while(cin>>n)
    {
        memset(visit,0,sizeof(visit));
        for(int i=0; i<n; ++i)
        {
            scanf("%lf%lf%lf",&a[i],&b[i],&c[i]);
            if(b[i] > c[i]) //预处理使起始时间均为0 
                b[i] += a[i];
            else
                b[i] -= a[i];
        }
        
        for(int i=0; i<n-1; ++i) //遍历每两个点是否相遇 
        {
            for(int j=i+1; j<n; ++j)
            {
                if(b[i] == b[j] && (b[i] > c[i] && b[j] > c[j] || b[i] < c[i] && b[j] < c[j]))  
                { //若起点相同且往同一个方向走
                    if(b[i] < c[i] && b[j] < c[j]) //同时往右走 
                    {
                        if( (b[i] + a[i] <= b[j]+a[j] && c[i] >= b[j]+a[j]) || (b[j] + a[j] <= b[i]+a[i] && c[j] >= b[i]+a[i]) )  
                        { //区间有交集 
                            visit[i]++;
                            visit[j]++;
                        }
                    }
                    else if(b[i] > c[i] && b[j] > c[j]) //同时往左走 
                    {
                        if( (b[i] - a[i] >= b[j] - a[j] && c[i] <= b[j] - a[j]) || (b[j] - a[j] >= b[i]-a[i] && c[j] <= b[i]-a[i]) )
                        { //区间有交集 
                            visit[i]++;
                            visit[j]++;
                        }
                    }
                }
                
                else
                {
                    double t = b[i] + b[j];
                    t /= 2; //求出相遇的点 
                    if(b[i] < c[i] && b[j] > c[j]) 
                    {
                        if(t >= b[i] + a[i] && t <= c[i] && t >= c[j] && t <= b[j] - a[j])  
                        { //交点在两区间内
                            visit[i]++;
                            visit[j]++;
                        }
                    }
                    else if(b[i] > c[i] && b[j] < c[j])
                    {
                        if(t <= b[i] - a[i] && t >= c[i] && t <= c[j] && t >= b[j] + a[j]) 
                        { //交点在两区间内 
                            visit[i]++;
                            visit[j]++;
                        }
                    }
                }
            }
        }
        for(int i=0; i<n; ++i)
            printf("%d%c",visit[i],i == n-1 ? '\n':' ');
    }
    return 0;
}
/*
2
1 2 3
2 2 1
*/

__EOF__

本文作者：特务依昂
本文链接：https://www.cnblogs.com/tuyang1129/p/9408497.html
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