HNU Knockout Racing

最新推荐文章于 2019-09-28 21:29:09 发布

转载最新推荐文章于 2019-09-28 21:29:09 发布 · 49 阅读

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原文链接：http://www.cnblogs.com/NaCl/p/9580206.html

本文针对一个特定的赛车位置查询问题，给出了详细的解析及C语言实现代码。问题描述为：在一条直道上，有n辆速度恒定的汽车往返于两点之间，需要回答m个询问，在指定时间内位于某一区间内的车辆数。

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题目链接http://acm.hnu.cn/online/?action=problem&type=show&id=13342&courseid=0
Problem description

The races became more popular than ever at Pandora planet. But these races are quite unusual. There are n cars participating in a race on the long straight track. Each car moves with a speed of 1 meter per second. Track has coordinates in meters.

The car number i moves between two points on the track with coordinates ai and bi starting at the second 0 in the point ai. The car moves from ai to bi, then from bi to ai, then from ai to bi again, and so on. Handsome Mike wants to knock some cars out of the race using dynamite. Thus he has m questions. The question number j is: what is the number of cars in the coordinates between xj and yj inclusive after tj seconds from the start? Your task is to answer Mike’s questions.

Input

The first line of the input file contains two integers n and m (1 ≤ n,m ≤ 1000) — the number of cars in the race and the number of questions. Each of the following n lines contains a description of the car: two integers ai and bi(0 ≤ ai,bi≤ 10^9, ai ≠ bi) — the coordinates of the two points between which the car i moves. Each of the following m lines contains a description of the question: three integers xj, yj, and tj (0 ≤ xj≤ yj≤ 10^9, 0 ≤ tj≤ 10^9) — the coordinate range and the time for the question j.

Output

Write m lines to the output file. Each line must contain one integer — the answer to the corresponding question in order they are given in the input file.

Sample Input
5 5
0 1
0 2
2 3
3 5
4 5
0 5 0
0 1 2
0 2 1
2 5 2
2 5 3

Sample Output
5
1
2
4
3

Problem Source
NEERC 2014

水题= =直接上代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define maxn 1000+100 
using namespace std;
struct node
{
    int x,y;
}a[maxn];
int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i=0;i<n;i++)
    {
        scanf("%d %d",&a[i].x,&a[i].y);
    }
    while(m--)
    {
        int xx,yy,t;
        scanf("%d %d %d",&xx,&yy,&t);
        int count=0;
        for(int i=0;i<n;i++)
        {
            int tx=0;
            if(t<(a[i].y-a[i].x))
            {
                tx=a[i].x+t;
            } 
            else
            {
                 int  u=t%(a[i].y-a[i].x);
                 int  v=t/(a[i].y-a[i].x); 
                 if(v&1) 
                 tx=a[i].y-u; 
                 else
                 tx=a[i].x+u; 
              } 
              if(tx>=xx&&tx<=yy)
              {
                  count++;
              } 
        }
        printf("%d\n",count);
    }
    return 0;
}