HDU_2579_bfs

http://acm.split.hdu.edu.cn/showproblem.php?pid=2579

简单bfs题，刚开始在纠结怎么存放vis，因为步数可能有几百步，这么多格子开数组的话也太多了，后来想到只要保存步数%k的状态就好了，bfs到达该点的步数肯定是最优的。

#include<iostream>
#include<string>
#include<cstring>
#include<queue>
using namespace std;

string s[];
int r,c,k,dir[][] = {-,,,-,,,,},vis[][][],flag;

struct point
{
    int x,y,counts;
}start;

int main()
{
    int n;
    cin >> n;
    while(n--)
    {
        flag = ;
        memset(vis,,sizeof(vis));
        cin >> r >> c >> k;
        for(int i = ;i < r;i++)   cin >> s[i];
        for(int i = ;i < r;i++)
        {
            for(int j = ;j < c;j++)
            {
                if(s[i][j] == 'Y')
                {
                    start.x = i;
                    start.y = j;
                    goto there;
                }
            }
        }
        there:
        start.counts = ;
        queue<point> q;
        q.push(start);
        vis[start.x][start.y][] = ;
        while(!q.empty())
        {
            int x = q.front().x,y = q.front().y,counts = q.front().counts;
            q.pop();
            if(s[x][y] == 'G')
            {
                flag = ;
                cout << counts << endl;
                break;
            }
            for(int i = ;i < ;i++)
            {
                int xx = x+dir[i][],yy = y+dir[i][],z = (counts+)%k;
                if(xx <  || xx >= r || yy <  || yy >= c)  continue;
                if(s[xx][yy] == '#' && z)  continue;
                if(vis[xx][yy][z])  continue;
                point temp;
                temp.x = xx;
                temp.y = yy;
                temp.counts = counts+;
                q.push(temp);
                vis[xx][yy][z] = ;
            }
        }
        if(!flag)   cout << "Please give me another chance!" << endl;
    }
    return ;
}