USACO The Tamworth Two

　这道题就是给你一张10*10的图和农夫和牛的位置， 农夫和牛刚开始往北走， 遇到障碍物或者边界会顺时针转向， 问你他们啥时候相遇。。不能相遇输出0.。直接将农夫和牛的位置和方向作为状态bfs即可。。代码如下：

/*
    ID: m1500293
    LANG: C++
    PROG: ttwo
*/

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>

using namespace std;
char Map[30][30];
int fx, fy, fdir;
int cx, cy, cdir;
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
bool inside(int x, int y) { return (x>=0&&x<10&&y>=0&&y<10); }
struct State
{
    int fx, fy, fdir;
    int cx, cy, cdir;
    int time;
    State() {}
    State(int fx1, int fy1, int fdir1, int cx1, int cy1, int cdir1, int time1)
    {
        fx = fx1; fy = fy1; fdir = fdir1;
        cx = cx1; cy = cy1; cdir = cdir1;
        time = time1;
    }
};

int vis[11][11][5][11][11][5];

int bfs()
{
    memset(vis, 0, sizeof(vis));
    queue<State> que;
    que.push(State(fx, fy, fdir, cx, cy, cdir, 0));
    vis[fx][fy][fdir][cx][cy][cdir] = 1;
    while(!que.empty())
    {
        State tp = que.front(); que.pop();
        if(tp.fx==tp.cx && tp.fy==tp.cy)
            return tp.time;
        int nfx=tp.fx+dx[tp.fdir], nfy=tp.fy+dy[tp.fdir], nfdir=tp.fdir;
        if(!inside(nfx, nfy) || Map[nfx][nfy]=='*')
        {
            nfx=tp.fx; nfy=tp.fy; nfdir=(tp.fdir+1)%4;
        }

        int ncx=tp.cx+dx[tp.cdir], ncy=tp.cy+dy[tp.cdir], ncdir=tp.cdir;
        if(!inside(ncx, ncy) || Map[ncx][ncy]=='*')
        {
            ncx=tp.cx; ncy=tp.cy; ncdir=(tp.cdir+1)%4;
        }
        if(vis[nfx][nfy][nfdir][ncx][ncy][ncdir]) return -1;            //发现重复
        que.push(State(nfx, nfy, nfdir, ncx, ncy, ncdir, tp.time+1));
        vis[nfx][nfy][nfdir][ncx][ncy][ncdir] = 1;
    }
    return -1;
}

int main()
{
    freopen("ttwo.in", "r", stdin);
    freopen("ttwo.out", "w", stdout);
    for(int i=0; i<10; i++)
    {
        scanf("%s", Map[i]);
        for(int j=0; j<10; j++)
        {
            if(Map[i][j] == 'F')
            {
                fx=i; fy=j; fdir=0;
                Map[i][j] = '.';
            }
            else if(Map[i][j] == 'C')
            {
                cx=i; cy=j; cdir=0;
                Map[i][j] = '.';
            }
        }
    }
    int res = bfs();
    if(res < 0)
        printf("0\n");
    else
        printf("%d\n", res);
    return 0;
}