Problem E: Erratic Ants

这个题没过……！
题意：小蚂蚁向四周走，让你在他走过的路中寻找最短路，其中可以反向
主要思路：建立想对应的图，寻找最短路径，其中错了好多次，到最后时间没过（1.没有考录反向2.没有考虑走过的路要标记……！！！！！内存超了……啊啊啊啊！！！！）
总之，这样了～～

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <map>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <cctype>

const double Pi = atan() * ;
using namespace std;

int graph[][];
bool through[][];
int dr[] = {,-,,};
int dc[] = {,,-,};
bool visit[][];
struct Point{
    int x,y;
    int step;
    Point(){
        step = ;
    }
    Point(int xx,int yy,int tt):x(xx),y(yy),step(tt){}
};
int main()
{
    freopen("input.in","r",stdin);
    //freopen("output.in","w",stdout);
    int t;
    cin >> t;
    queue<Point>que;
    while(t--){
        int n;
        cin >> n;
        memset(graph,,sizeof(graph));
        memset(through,,sizeof(through));
        memset(visit,,sizeof(visit));
        int x = ;
        int y = ;
        int sx = ;
        int sy = ;
        graph[x][y] = ;
        char ch;
        int ww = n;
        while(n--){
            cin >> ch;
            int xx = x;
            int yy = y;
            if(ch == 'E'){
                xx++;
            }
            else if(ch == 'W'){
                xx--;
            }
            else if(ch == 'S'){
                yy--;
            }
            else if(ch == 'N'){
                yy++;
            }
            if(!graph[xx][yy])
                graph[xx][yy] = graph[x][y]+;
            through[ graph[x][y] ][graph[xx][yy] ] = ;
            through[ graph[xx][yy] ][graph[x][y] ] = ;
            x = xx;
            y = yy;
        }
        if(ww == ){
            cout << "" << endl;
            continue;
        }
        while(!que.empty()){
            que.pop();
        }
        Point head(sx,sy,);
        que.push(head);
        visit[sx][sy] = ;
        while(!que.empty()){
            Point tmp = que.front();
            que.pop();
            if(tmp.x == x && tmp.y == y){
                cout << tmp.step << endl;
                break;
            }
            for(int i = ;i < ;i++){
                int xx = tmp.x + dr[i];
                int yy = tmp.y + dc[i];
                if(graph[xx][yy] !=  && through[  graph[tmp.x][tmp.y] ][ graph[xx][yy] ] && !visit[xx][yy]){
                    Point tt(xx,yy,tmp.step+);
                    que.push(tt);
                    visit[xx][yy] = ;
                }
            }
        }
    }
    return ;
}