[洛谷] P2689 东南西北
求个曼哈顿距离然后贪心
//#pragma GCC optimize(2)
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <string>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <ctime>
#include <vector>
#include <fstream>
#include <list>
#include <iomanip>
#include <numeric>
using namespace std;
typedef long long ll;
const int MAXN = 1e6 + 10;
int arr[MAXN][2], bx, by, ex, ey, step = 0, n;
bool judge(int tx, int ty, int x, int y)
{
double tdis = sqrt((tx - ex) * (tx - ex) +(ty - ey) * (ty - ey));
double rdis = sqrt((x - ex) * (x - ex) +(y - ey) * (y - ey));
if(tdis < rdis)
return true;
return false;
}
int main()
{
//ios::sync_with_stdio(false);
//cin.tie(0); cout.tie(0);
cin>>bx>>by>>ex>>ey>>n;
char c;
for(int i = 0; i < n; i++)
{
cin>>c;
switch(c)
{
case 'E':
arr[i][0] = 1, arr[i][1] = 0;
break;
case 'S':
arr[i][0] = 0, arr[i][1] = -1;
break;
case 'W':
arr[i][0] = -1, arr[i][1] = 0;
break;
case 'N':
arr[i][0] = 0, arr[i][1] = 1;
break;
}
}
int tx, ty;
for(int i = 0; i < n; i++)
{
tx = bx + arr[i][0], ty = by + arr[i][1];
if(judge(tx, ty, bx, by))
bx = tx, by = ty, step++;
if(bx == ex && by == ey)
{
cout<<step<<endl;
goto l1;
}
}
cout<<"-1"<<endl;
l1:
return 0;
}