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来源：互联网发布：怎么恢复u盘数据编辑：程序博客网时间：2024/05/08 12:45
#include <stdio.h>#include <iostream>#include <cstring>#include <stdlib.h>#include <algorithm>#include <queue>#include <math.h>#define MAX 300#define INF 1e8using namespace std;typedef pair<int,int> P;int N,M;char maze[MAX][MAX];int yx,yy;//y的起始坐标int mx,my;//m的起始坐标int d1[MAX][MAX],d2[MAX][MAX];int move1[4] = {1,0,-1,0},move2[4] = {0,1,0,-1};int ax[40005],ay[40005];void dfs(int x,int y,int x1,int y1){    queue<P> que;    P b;    que.push(P(x,y));    for (int i = 0;i < N;i++)    {        for (int j = 0;j < M;j++)        {            d1[i][j] = INF;            d2[i][j] = INF;        }    }    d1[x][y] = 0;    d2[x1][y1] = 0;    while (!que.empty())    {        b = que.front();        que.pop();        int nx,ny;        for (int i = 0;i < 4;i++)        {            ny = b.second + move1[i];            nx = b.first +move2[i];            if((nx>=0&&nx<N)&&(ny>=0&&ny<M)&&maze[nx][ny]!='#'&&d1[nx][ny]==INF)            {                que.push(P(nx,ny));                d1[nx][ny] = d1[b.first][b.second] + 1;            }        }    }    que.push(P(x1,y1));    while (!que.empty())    {        b = que.front();        que.pop();        int nx,ny;        for (int i = 0;i < 4;i++)        {            ny = b.second + move1[i];            nx = b.first +move2[i];            if((nx>=0&&nx<N)&&(ny>=0&&ny<M)&&maze[nx][ny]!='#'&&d2[nx][ny]==INF)            {                que.push(P(nx,ny));                d2[nx][ny] = d2[b.first][b.second] + 1;            }        }    }}int main(void){    int k;    int ans;    while(scanf("%d %d",&N,&M) != EOF)    {        if(N==0&&M==0)        {            break;        }        k = 0;        for (int i = 0;i < N;i++)        {            scanf("%s",maze[i]);        }        for (int i = 0;i < N;i++)        {            for (int j = 0;j < M;j++)            {                if(maze[i][j] == 'Y')                    yx = i,yy = j;                else if(maze[i][j] == 'M')                    mx = i,my = j;                else if(maze[i][j] == '@')                {                    ax[k] = i,ay[k] = j;                    k++;                }            }        }        dfs(yx,yy,mx,my);        ans = d1[ax[0]][ay[0]] + d2[ax[0]][ay[0]];        for (int i = 1;i < k;i++)        {            ans = min(ans,d1[ax[i]][ay[i]] + d2[ax[i]][ay[i]]);            /*cout<<d2[ay[i]][ax[i]]<<endl;            cout<<d1[ay[i]][ax[i]]<<endl;*/        }        cout<<ans*11<<endl;    }    return 0;}
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