【Codeforces 746 C Tram】+ 细节

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我是靠谱客的博主诚心手套，最近开发中收集的这篇文章主要介绍【Codeforces 746 C Tram】+ 细节，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

C. Tram
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

The tram in Berland goes along a straight line from the point 0 to the point s and back, passing 1 meter per t1 seconds in both directions. It means that the tram is always in the state of uniform rectilinear motion, instantly turning around at points x = 0 and x = s.

Igor is at the point x1. He should reach the point x2. Igor passes 1 meter per t2 seconds.

Your task is to determine the minimum time Igor needs to get from the point x1 to the point x2, if it is known where the tram is and in what direction it goes at the moment Igor comes to the point x1.

Igor can enter the tram unlimited number of times at any moment when his and the tram’s positions coincide. It is not obligatory that points in which Igor enter and exit the tram are integers. Assume that any boarding and unboarding happens instantly. Igor can move arbitrary along the line (but not faster than 1 meter per t2 seconds). He can also stand at some point for some time.
Input

The first line contains three integers s, x1 and x2 (2 ≤ s ≤ 1000, 0 ≤ x1, x2 ≤ s, x1 ≠ x2) — the maximum coordinate of the point to which the tram goes, the point Igor is at, and the point he should come to.

The second line contains two integers t1 and t2 (1 ≤ t1, t2 ≤ 1000) — the time in seconds in which the tram passes 1 meter and the time in seconds in which Igor passes 1 meter.

The third line contains two integers p and d (1 ≤ p ≤ s - 1, d is either 1 or ) — the position of the tram in the moment Igor came to the point x1 and the direction of the tram at this moment. If , the tram goes in the direction from the point s to the point 0. If d = 1, the tram goes in the direction from the point 0 to the point s.
Output

Print the minimum time in seconds which Igor needs to get from the point x1 to the point x2.
Examples
Input

4 2 4
3 4
1 1

Output

Input

5 4 0
1 2
3 1

Output

Note

In the first example it is profitable for Igor to go by foot and not to wait the tram. Thus, he has to pass 2 meters and it takes 8 seconds in total, because he passes 1 meter per 4 seconds.

In the second example Igor can, for example, go towards the point x2 and get to the point 1 in 6 seconds (because he has to pass 3 meters, but he passes 1 meters per 2 seconds). At that moment the tram will be at the point 1, so Igor can enter the tram and pass 1 meter in 1 second. Thus, Igor will reach the point x2 in 7 seconds in total.

题意：给出轨道长度 s，初始位置 x1，目标位置 x2，轨车速度 t1/1m， Igor 速度 t2 / 1m，轨车位置 p ，前进方向 D，1表示 0 -> s, -1 表示 s -> 0~轨车走到头改变方向～求从x1到x2所发费的最少时间

思路：模拟轨车的前进方向以及p，x1，x2的位置关系，都是可以坐上车的，然后与直接步行的时间比较～

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f;
int t[100];
int main()
{
int s,x1,x2,t1,t2,p,d;
scanf("%d %d %d %d %d %d %d",&s,&x1,&x2,&t2,&t1,&p,&d);
t[1] = abs(x1 - x2) * t1;
for(int i = 2 ; i <= 11 ; i++)
t[i] = INF;
if(d > 0){
if(x2 > p){
if(x1 >= p){
if(x1 <= x2) t[2] = (x2 - p) * t2;
else t[3] = (2 * s - x2 - p) * t2;
}
else t[4] = (s - p + s + x2) * t2;
}
else{
if(x1 >= x2) t[5] = (s - x2 + s - p) * t2;
else t[6] = (s - p + s + x2) * t2;
}
}
else{
if(x2 < p){
if(x1 <= p){
if(x1 >= x2) t[7] = (p - x2) * t2;
else t[8] = (p + x2) * t2;
}
else
t[9] = (p + s + s - x2) * t2;
}
else{
if(x1 <= x2) t[10] = (p + x2) * t2;
else t[11] = (p + s + s - x2) * t2;
}
}
int ans = t[1];
for(int i = 2 ; i <= 11 ; i++)
ans = min(ans,t[i]);
printf("%dn",ans);
return 0;
}