Codeforces Round #180 (Div. 2) B. Sail 贪心

B. Sail

题目连接：

http://www.codeforces.com/contest/298/problem/B

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

5 0 0 1 1

SESNW

Sample Output

4

Hint

题意

给你起点sx,sy，终点ex,ey

然后给你n个指令，指令你可以选择遵守也可以选择不遵守

问你最短多久可以从起点到达终点

题解：

贪心，如果我选择之后，能够离终点更近一点，那么我就会选择，否则我就不会选择。

代码

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int t;
    long long sx,sy,ex,ey;
    cin>>t>>sx>>sy>>ex>>ey;
    if(sx==ex&&sy==ey)return puts("0");
    string s;cin>>s;
    for(int i=0;i<t;i++)
    {
        if(s[i]=='E'&&sx<ex)sx+=1;
        if(s[i]=='W'&&sx>ex)sx-=1;
        if(s[i]=='N'&&sy<ey)sy+=1;
        if(s[i]=='S'&&sy>ey)sy-=1;
        if(sx==ex&&sy==ey)
        {
            printf("%d",i+1);
            return 0;
        }
    }
    return puts("-1");
}