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来源：互联网发布：windows触摸屏无法使用编辑：程序博客网时间：2024/05/22 17:03
OJ题目：点~~
const int Max_N = 25;const int inf = 100000000;int N , M;char table[Max_N][Max_N];int dir[4][2] = {{-1,0},{0,1},{1,0},{0,-1}};int minstep[Max_N][Max_N][Max_N][Max_N];struct Point//表示Ball的结构体{    int x;    int y;    int OK;//1表示已搞定 ， 0表示还没搞定    Point(){}    Point(int i , int j , int k):x(i),y(j),OK(k){}};struct Node//表示状态的结构体{    Point A;    Point B;    int step;//本状态的步数    Node(){}    Node(Point i , Point j , int s):A(i),B(j),step(s){}};vector<Point> List_Ball;int bfs(Node s){    int i , j , k , t;    for(i = 1;i <= N;i++)        for(j = 1;j <= M;j++)            for(k = 1;k <= N;k++)                for(t = 1;t <= M;t++)                    minstep[i][j][k][t] = inf;    queue<Node> que;    que.push(s);    Node now ;    Point newA , newB;    while(!que.empty())//bfs    {        now  = que.front();        que.pop();        if(now.A.OK&& now.B.OK)//如果两个Ball都已经进Hole ， 成功 ， 返回步数            return now.step;        int nextAx , nextBx , nextAy , nextBy;        for(i = 0;i < 4;i++)        {            nextAx = now.A.x + dir[i][0];            nextAy = now.A.y + dir[i][1];            nextBx = now.B.x + dir[i][0];            nextBy = now.B.y + dir[i][1];            if(table[nextAx][nextAy] == '*' && table[nextBx][nextBy] == '*')//如果下面两个位置都是墙，则跳过                continue;            if(now.A.OK||table[nextAx][nextAy] == '*')//如果A已搞定，或者下个位置是墙，A的位置不变            {                nextAx = now.A.x;                nextAy = now.A.y;            }            if(now.B.OK ||table[nextBx][nextBy] == '*')//如果B已搞定，或者下个位置是墙，B的位置不变            {                nextBx = now.B.x;                nextBy = now.B.y;            }            if(!now.A.OK && !now.B.OK && nextAx == nextBx && nextAy == nextBy)//如果AB都在，且下个位置重合，则跳过                continue;            newA.x = nextAx;            newA.y = nextAy;            newB.x = nextBx;            newB.y = nextBy;            if(now.A.OK)//如果A已经搞定，下面考虑B            {                newA.OK = 1;//A的状态要赋值！！                if(table[nextBx][nextBy] == 'H'&& !(nextAx == nextBx && nextAy == nextBy))//如果B的新位置刚好是Hole，且不是A的那个Hole，则B搞定                    newB.OK = 1;                else                    newB.OK = 0;            }            else if(now.B.OK)//如果B已经搞定，下面考虑A                {                    newB.OK = 1;//B的状态要赋值！！                    if(table[nextAx][nextAy] == 'H' && !(nextBx == nextAx && nextBy == nextAy))//如果A的新位置刚好是Hole，且不是B的那个Hole，则A搞定                        newA.OK = 1;                    else                        newA.OK = 0;                }                else//如果A ,B都未搞定                {                    if(table[nextAx][nextAy] == 'H')//考虑A                    {                        newA.OK = 1;                    }                    else                        newA.OK = 0;                    if(table[nextBx][nextBy] == 'H')//考虑B                    {                        newB.OK = 1;                    }                    else                        newB.OK = 0;                }            if(minstep[newA.x][newA.y][newB.x][newB.y] > now.step + 1)//如果新状态步数比之前的少，则更新            {                minstep[newA.x][newA.y][newB.x][newB.y] = now.step + 1;                que.push(Node(newA , newB , now.step + 1));            }        }    }    return -1;}int main(){    int Case;    cin >> Case;    while(Case--)    {        scanf("%d%d" , &N ,&M);        List_Ball.clear();        int i , j;        for(i = 1;i <= N;i++)            scanf("%s",table[i]+1);        for(i = 1;i <= N;i++)            for(j = 1;j <= M;j++)            if(table[i][j] == 'B')                List_Ball.push_back(Point(i , j , 0));//List_Ball保存两个Ball        int ans = bfs(Node(List_Ball[0] , List_Ball[1] , 0));//广搜        if(ans == -1) cout << "Sorry , sir , my poor program fails to get an answer." << endl ;        else            cout << ans << endl;    }    return 0;}
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