codeforce 298 B Sail
B. Sail
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int sx,sy,ex,ey;
int t;
char direc[100100];
int nneed,eneed,wneed,sneed;//每次都得初始化
int n,e,w,s;//每次都得初始化
void caculate()
{
nneed=0;eneed=0;sneed=0;wneed=0;
int tempx=ex-sx;
int tempy=ey-sy;
if(tempx>=0)
{
eneed=tempx;
}
else
{
wneed=-tempx;
}
if(tempy>=0)
{
nneed=tempy;
}
else
{
sneed=-tempy;
}
}
int isposi(int a)
{
if(a<0)
return 0;
else return a;
}
int compare()
{
if(isposi(nneed-n)==0&&isposi(sneed-s)==0&&isposi(eneed-e)==0&&isposi(wneed-w)==0)
return 1;
else return 0;
}
void solve()
{
n=0;s=0;e=0;w=0;
scanf("%s",direc);
for(int i=0;i<t;i++)
{
switch(direc[i])
{
case 'N':
{
n++;
break;
}
case 'S':
{
s++;
break;
}
case 'E':
{
e++;
break;
}
case 'W':
{
w++;
break;
}
}
if(compare()==1)
{
printf("%d\n",i+1);
return;
}
}
printf("-1\n");
}
int main(int argc, char *argv[])
{
while(scanf("%d %d %d %d %d",&t,&sx,&sy,&ex,&ey)!=EOF)
{
caculate();//计算需要的n，e，s，w
solve();
}
return 0;
}