HDU-3085-Nightmare Ⅱ(双向BFS)

浏览数：33 / 时间：2015年06月09日

Problem Description

Last night, little erriyue had a horrible nightmare. He dreamed that he and his girl friend were trapped in a big maze separately. More terribly, there are two ghosts in the maze. They will kill the people. Now little erriyue wants to know if he could find his girl friend before the ghosts find them.
You may suppose that little erriyue and his girl friend can move in 4 directions. In each second, little erriyue can move 3 steps and his girl friend can move 1 step. The ghosts are evil, every second they will divide into several parts to occupy the grids within 2 steps to them until they occupy the whole maze. You can suppose that at every second the ghosts divide firstly then the little erriyue and his girl friend start to move, and if little erriyue or his girl friend arrive at a grid with a ghost, they will die.
Note: the new ghosts also can devide as the original ghost.

Input

The input starts with an integer T, means the number of test cases.
Each test case starts with a line contains two integers n and m, means the size of the maze. (1<n, m<800)
The next n lines describe the maze. Each line contains m characters. The characters may be:
‘.’ denotes an empty place, all can walk on.
‘X’ denotes a wall, only people can’t walk on.
‘M’ denotes little erriyue
‘G’ denotes the girl friend.
‘Z’ denotes the ghosts.
It is guaranteed that will contain exactly one letter M, one letter G and two letters Z.

Output

Output a single integer S in one line, denotes erriyue and his girlfriend will meet in the minimum time S if they can meet successfully, or output -1 denotes they failed to meet.

Sample Input

3
5 6
XXXXXX
XZ..ZX
XXXXXX
M.G...
......
5 6
XXXXXX
XZZ..X
XXXXXX
M.....
..G...

10 10
..........
..X.......
..M.X...X.
X.........
.X..X.X.X.
.........X
..XX....X.
X....G...X
...ZX.X...
...Z..X..X

Sample Output

1
1
-1

Author

二日月

Source

HDU 2nd “Vegetable-Birds Cup” Programming Open Contest

思路：鬼分身所在的地方就直接变成墙，接着女孩走1步，男孩走3步，相遇就直接返回时间。(注意：鬼的分身也是可以继续分的)

#include <stdio.h>
#include <string.h>

struct{
int x,y,step;
}que1[1000000],que2[1000000],que[1000000],t;

int n,m,gxa,gya,gxb,gyb,xa,ya,xb,yb,nxt[4][2]={{0,1},{1,0},{0,-1},{-1,0}},top,bottom,gt;
char mp[800][805];
bool vis1[800][800],vis2[800][800],vis[800][800];

void divide()//鬼分身，让有鬼的地方都变成墙
{
    int i;

    while(top<bottom)
    {
        t=que[top];

        if(t.step>=gt) return;

        t.step++;

        for(i=0;i<4;i++)
        {
            t.x+=nxt[i][0];
            t.y+=nxt[i][1];

            if(t.x>=0 && t.x<n && t.y>=0 && t.y<m && !vis[t.x][t.y])
            {
                vis[t.x][t.y]=1;

                mp[t.x][t.y]='X';

                que[bottom++]=t;
            }

            t.x-=nxt[i][0];
            t.y-=nxt[i][1];
        }

        top++;
    }
}

int bfs()
{
    int i,t1,t2,top1,top2,bottom1,bottom2;

    memset(vis,0,sizeof vis);
    memset(vis1,0,sizeof vis1);
    memset(vis2,0,sizeof vis2);

    top=0;
    bottom=2;
    que[0].x=gxa;
    que[0].y=gya;
    que[0].step=0;
    que[1].x=gxb;
    que[1].y=gyb;
    que[1].step=0;

    gt=2;
    t1=1;
    t2=3;

    top1=0;
    bottom1=1;
    que1[0].x=xa;
    que1[0].y=ya;
    que1[0].step=0;
    vis1[xa][ya]=1;

    top2=0;
    bottom2=1;
    que2[0].x=xb;
    que2[0].y=yb;
    que2[0].step=0;
    vis2[xb][yb]=1;

    while(1)
    {
        divide();//鬼先分身

        bool flag=0;

        while(top1<bottom1)//女孩走
        {
            t=que1[top1];

            if(t.step>=t1) break;

            t.step++;

            for(i=0;i<4;i++)
            {
                t.x+=nxt[i][0];
                t.y+=nxt[i][1];

                if(t.x>=0 && t.x<n && t.y>=0 && t.y<m && mp[t.x-nxt[i][0]][t.y-nxt[i][1]]!='X' && mp[t.x][t.y]!='X' && !vis1[t.x][t.y])
                {
                    flag=1;

                    if(vis2[t.x][t.y]) return t1;

                    vis1[t.x][t.y]=1;
                    que1[bottom1++]=t;
                }

                t.x-=nxt[i][0];
                t.y-=nxt[i][1];
            }

            top1++;
        }

        while(top2<bottom2)//男孩走
        {
            t=que2[top2];

            if(t.step>=t2) break;

            t.step++;

            for(i=0;i<4;i++)
            {
                t.x+=nxt[i][0];
                t.y+=nxt[i][1];

                if(t.x>=0 && t.x<n && t.y>=0 && t.y<m && mp[t.x-nxt[i][0]][t.y-nxt[i][1]]!='X' && mp[t.x][t.y]!='X' && !vis2[t.x][t.y])
                {
                    flag=1;

                    if(vis1[t.x][t.y]) return t1;

                    vis2[t.x][t.y]=1;
                    que2[bottom2++]=t;
                }

                t.x-=nxt[i][0];
                t.y-=nxt[i][1];
            }

            top2++;
        }

        if(!flag) return -1;

        gt+=2;
        t1++;
        t2+=3;
    }
}

int main()
{
    int T,i,j,cnt;

    scanf("%d",&T);

    while(T--)
    {
        scanf("%d%d",&n,&m);

        for(i=0;i<n;i++) scanf("%s",mp[i]);

        cnt=0;

        for(i=0;i<n;i++) for(j=0;j<m;j++)
        {
            if(mp[i][j]=='G') xa=i,ya=j;
            else if(mp[i][j]=='M') xb=i,yb=j;
            else if(mp[i][j]=='Z')
            {
                mp[i][j]='X';

                if(!cnt) gxa=i,gya=j;
                else gxb=i,gyb=j;

                cnt++;
            }
        }

        printf("%d\n",bfs());
    }
}

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HDU-3085-Nightmare Ⅱ(双向BFS)