Sail

Description
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 Input
Input
5 0 0 1 1
SESNW
Output
4
Input
10 5 3 3 6
NENSWESNEE
Output
-1

题目意思：

一开始你的船在(sx,sy)处，要去(ex.ey)，给定t秒内每一秒的风向，每一秒可以选择顺着风的方向走一格或者不动，问能否在t秒内到达目的地，可以的话最早可以在多少秒到。

解题思路：

只有当顺着风的方向可以使当前位置到目的地距离减小，否则就不动，如果在中间某一时刻到目的地了就输出当时的秒数，如果到最后都没到就是到不了了。

上代码：

#include<stdio.h>
#include<string.h>
int main()
{
    int t,sx,sy,ex,ey,i;
    char s[1000001];
    scanf("%d%d%d%d%d",&t,&sx,&sy,&ex,&ey);
    getchar();
    gets(s);
    for(i=0;i<t;i++)
    {
        if(s[i]==‘E‘&&sx<ex)
            sx++;
        else  if(s[i]==‘W‘&&sx>ex)
            sx--;
        else if(s[i]==‘S‘&&sy>ey)
            sy--;
        else if(s[i]==‘N‘&&sy<ey)
            sy++;
        if(sx==ex&&sy==ey)
        {
            printf("%d\n",i+1);
            break;
        }
    }
    if(!(sx==ex&&sy==ey))
        printf("-1\n");
    return 0;
}