Traffic Jam in Flower Town
题目链接：
http://acm.hust.edu.cn/vjudge/contest/126546#problem/I
Description
Having returned from Sun City, Dunno told all his friends that every shorty may have a personal
automobile. Immediately after that so many citizens took a fancy of becoming road-users, that Bendum
and Twistum had to make a mass production of cars on soda water with syrup. Now traffic jams from
several cars occasionally appear on the crossing of Bell-flower Street and Daisy Street.
Bell-flower Street goes from the South to the North and has two driving paths. It has the right driving,
i. e. automobiles move from the South to the North on the Eastern path and from the North to the South
on the Western path. Daisy Street is single-pathed, and it is perpendicular to Bell-flower Street. There is
one-way movement on it, but its driving direction is organized in such a way that automobiles drive away
from the crossroad in two opposite directions (see the picture).
Yesterday on his way home Dunno saw cars standing in a traffic jam
on Bell-flower Street from different sides of the crossing with Daisy
Street. Some of the drivers wanted to go forward, some wanted to turn
right or left. An automobile can pass the crossing in one second, but if
the driver is turning left, he first have to let pass all oncoming vehicles,
going forward and to the right. How many seconds did it take all the
cars to pass the crossing, providing that no other cars drove up to the
crossing?
Input
The first line contains the sequence of symbols “F”, “L” and “R”, describing directions in which drivers who
arrived to the crossing from the South wanted to go. “F” stands for those drivers who were going forward,
“L” is for those who were turning left, and “R” is for those who were turning right. Automobiles are listed
in the order from the closest to the crossing to the farthest one. The second line contains the description
of the cars, arrived to the crossing from the North, in the same form. Both sequences have length from 1
to 1 000.
Output
Output time in seconds, which took all the cars to pass the crossing.
Examples
RLF
FF
4
L
L
1
Explanation
In the first example we number the cars from 1 to 5 in the order described in the input data. Then
in the first second the crossing was passed by the first and the fourth cars because they didn’t cause
an obstruction to each other. Then the second car was turning left and had to let the fifth car pass. As
a result, at each of the following three seconds only one car passed the crossing, and their order was as
follows: the fifth one, the second one and the third one.
In the second example the cars didn’t cause any obstruction to each other and turned simultaneously.
##题意：
如图所示的道路通行方向.
左转时要先让对面的直行和右转先走.
给出两个方向来的车辆将要走的方向.
给出总共需要的时间. (不冲突的方向可以同时走).
##题解：
直接模拟一遍就可以了.
这题读题比做题要难呀.
队友写的代码，我就不重复写了，挂上来做个记录.
##代码：
``` cpp
#include
#include
#include
using namespace std;
char s1[1005], s2[1005];
int main() {
gets(s1); gets(s2);
int len1 = strlen(s1);
int len2 = strlen(s2);
int pos1 = 0, pos2 = 0;
int ans = 0;
while (1) {
if (pos1 == len1 && pos2 == len2) break;
if (pos1 == len1) {ans++; pos2++; continue;}
if (pos2 == len2) {ans++; pos1++; continue;}
if (s1[pos1] == 'F' && s2[pos2] == 'F') {
pos1++, pos2++; ans++; continue;
}
if (s1[pos1] == 'R' && s2[pos2] == 'R') {
pos1++, pos2++; ans++; continue;
}
if (s1[pos1] == 'R' && s2[pos2] == 'F') {
pos1++, pos2++; ans++; continue;
}
if (s1[pos1] == 'F' && s2[pos2] == 'R') {
pos1++, pos2++; ans++; continue;
}
if (s1[pos1] == 'L' && s2[pos2] == 'L') {
pos1++, pos2++; ans++; continue;
}
if (s1[pos1] == 'L') {
pos2++; ans++; continue;
}
if (s2[pos2] == 'L') {
pos1++; ans++; continue;
}
}
printf("%d\n", ans);
return 0;
}