Codeforces-298b H Sail

B. Sail

time limit per test
1 second

memory limit per test
256 megabytes

The polar bears are going fishing. They plan to sail from
(sx, sy) to
(ex, ey). However, the boat can only sail by wind. At each second, the wind blows in one of these directions: east, south, west or
north. Assume the boat is currently at (x, y).

If the wind blows to the east, the boat will move to (x + 1, y).

If the wind blows to the south, the boat will move to
(x, y - 1).
If the wind blows to the west, the boat will move to (x - 1, y).

If the wind blows to the north, the boat will move to
(x, y + 1).
Alternatively, they can hold the boat by the anchor. In this case, the boat stays at
(x, y). Given the wind direction for
t seconds, what is the earliest time they sail to
(ex, ey)?

Input
The first line contains five integers t, sx, sy, ex, ey
(1 ≤ t ≤ 105,  - 109 ≤ sx, sy, ex, ey ≤ 109).
The starting location and the ending location will be different.

The second line contains t characters, the
i-th character is the wind blowing direction at the
i-th second. It will be one of the four possibilities: "E" (east), "S" (south), "W" (west) and "N" (north).

Output
If they can reach (ex, ey) within
t seconds, print the earliest time they can achieve it. Otherwise, print "-1" (without quotes).

Sample test(s)

Input

5 0 0 1 1
SESNW

Output

4

Input

10 5 3 3 6
NENSWESNEE

Output

-1

Note
In the first sample, they can stay at seconds 1,
3, and move at seconds 2,
4.

In the second sample, they cannot sail to the destination.

——————————————————亚历山大的分割线——————————————————

思路：上高中的时候，数学老师教过一种小明从家里去学校有多少种路径的排列组合。。。咳咳，本题跟排列组合是木有关系滴！但是这个想法是要有的。横坐标差值对应了N或者S，纵坐标差值对应了W或者E。当前风向并不是期望的风向之一的话，等待一秒。

代码如下：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char dir[100010];
char c, d;
int sx, sy, ex, ey;
void jud(){//jud()函数用来储存期望的风向，c表示东西，d表示南北
if(ex > sx&&ey > sy)
{d = 'N'; c = 'E';}
else if(ex < sx&&ey > sy)
{d = 'N'; c = 'W';}
else if(ex < sx&&ey < sy)
{d = 'S'; c = 'W';}
else if(ex > sx&&ey < sy)
{d = 'S'; c = 'E';}
else if(sx == ex)
{d = ey > sy ? 'N' : 'S'; c = ' ';}
else
{c = ex > sx ? 'E' : 'W'; d = ' ';}
}
int main(){
int T, i, cou = 0;
int stx, sty, step;
scanf("%d%d%d%d%d", &T, &sx, &sy, &ex, &ey);
scanf("%s", dir);
stx = abs(sx - ex);//横坐标差值
sty = abs(sy - ey);//纵坐标差值
step = stx + sty;//一共需要的最少步数
jud();
for(i = 0; i < T; i++){
if(dir[i] != c&&dir[i] != d) cou++;//cou的大小表示需要等待的时间
  else if(dir[i] == c){
if(stx) stx--;
else cou++;
}
else if(dir[i] == d){
if(sty) sty--;
else cou++;
}
if(!stx && !sty) break;
}
if(stx || sty)
printf("-1\n");
else
printf("%d\n", cou + step);
return 0;
}