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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

題目意思：

一開始你的船在(sx,sy)處，要去(ex.ey)，給定t秒內每一秒的風向，每一秒可以選擇順著風的方向走一格或者不動，問能否在t秒內到達目的地，可以的話最早可以在多少秒到。

解題思路：

只有當順著風的方向可以使當前位置到目的地距離減小，否則就不動，如果在中間某一時刻到目的地了就輸出當時的秒數，如果到最後都沒到就是到不了了。

上代碼：

#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;
}

Sail