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(1) * 4;

using namespace std;



int graph[200][200];

bool through[100][100];

int dr[] = {1,-1,0,0};

int dc[] = {0,0,-1,1};

bool visit[200][200];

struct Point{

    int x,y;

    int step;

    Point(){

        step = 0;

    }

    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,0,sizeof(graph));

        memset(through,0,sizeof(through));

        memset(visit,0,sizeof(visit));

        int x = 100;

        int y = 100;

        int sx = 100;

        int sy = 100;

        graph[x][y] = 1;

        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]+1;

            through[ graph[x][y] ][graph[xx][yy] ] = 1;

            through[ graph[xx][yy] ][graph[x][y] ] = 1;

            x = xx;

            y = yy;

        }

        if(ww == 1){

            cout << "1" << endl;

            continue;

        }

        while(!que.empty()){

            que.pop();

        }

        Point head(sx,sy,0);

        que.push(head);

        visit[sx][sy] = 1;

        while(!que.empty()){

            Point tmp = que.front();

            que.pop();

            if(tmp.x == x && tmp.y == y){

                cout << tmp.step << endl;

                break;

            }

            for(int i = 0;i < 4;i++){

                int xx = tmp.x + dr[i];

                int yy = tmp.y + dc[i];

                if(graph[xx][yy] != 0 && through[  graph[tmp.x][tmp.y] ][ graph[xx][yy] ] && !visit[xx][yy]){

                    Point tt(xx,yy,tmp.step+1);

                    que.push(tt);

                    visit[xx][yy] = 1;

                }

            }

        }

    }

    return 0;

}

View Code