CodeForces #589 D Boulevard 模拟

题目描述：

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

The boulevard can be represented as the axis Ox. For every person there are three parameters characterizing the behavior: ti, si, fi — 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 si to fi 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 si she immediately starts walking towards the point fi.

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 si and fi. After a person achieves the destination point fi she 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 ti, si, fi(1 ≤ ti, si, fi ≤ 106,  si ≠ fi), where ti, si, fi — 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, …, rn separated by a space, where ri denotes the number which the i-th person greets other people while walking along the boulevard.

Sample Input
Input

31 1 105 8 29 9 10

Output

2 1 1 

Input

33 2 44 3 43 6 4

Output

2 2 2 

题目分析：

有n个人，分别在时间ti的时候从si出发，走向ei。每个人在走的过程中会遇到别的人，将他们遇到别人的个数分别输出。
模拟。

代码如下：

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>using namespace std;struct node{    int s,e;    int t;} p[1010];int n;int ans[1010];bool judge(int a,int b,int c,int d)//两人同方向时，时间是否能重合{    if(c<=a && b<=d) return 1;    if(a<=c && c<=b) return 1;    if(a<=d && d<=b) return 1;    return 0;}int main(){    while(scanf("%d",&n)!=-1)    {        memset(ans,0,sizeof(ans));        for(int i=1; i<=n; i++)        {            scanf("%d%d%d",&p[i].t,&p[i].s,&p[i].e);        }        for(int i=1; i<=n; i++)        {            for(int j=i+1; j<=n; j++)            {                if (p[i].s<=p[i].e && p[j].s<=p[j].e)//从左向右                {                    if(p[i].s-p[i].t==p[j].s-p[j].t && judge(p[i].t,p[i].t+p[i].e-p[i].s,p[j].t,p[j].t+p[j].e-p[j].s))                        ans[i]++,ans[j]++;                }                if (p[i].s<=p[i].e && p[j].s>p[j].e)//从左向右和从右向左                {                    double t=p[i].t-p[i].s+p[j].t+p[j].s;                    t/=2;//两人相遇点                    if(t>=p[i].t && t<=p[i].t+p[i].e-p[i].s && t>=p[j].t && t<=p[j].t+p[j].s-p[j].e)                        ans[i]++,ans[j]++;                }                if (p[i].s>p[i].e && p[j].s<=p[j].e)//从右向左和从左向右                {                    double t=p[i].t+p[i].s+p[j].t-p[j].s;                    t/=2;//两人相遇点                    if(t>=p[j].t && t<=p[j].t+p[j].e-p[j].s && t>=p[i].t && t<=p[i].t+p[i].s-p[i].e)                     ans[i]++,ans[j]++;                }                if (p[i].s>p[i].e && p[j].s>p[j].e)//从右向左                {                    if(p[i].s+p[i].t==p[j].s+p[j].t && judge(p[i].t,p[i].t+p[i].s-p[i].e,p[j].t,p[j].t+p[j].s-p[j].e))                     ans[i]++,ans[j]++;                }            }        }        printf("%d",ans[1]);        for(int i=2; i<=n; i++)        {            printf(" %d",ans[i]);        }        printf("\n");    }    return 0;}