HDU-2612-Find a way

来源：互联网发布：java高斯模糊编辑：程序博客网时间：2024/04/29 21:54

//这题坑了我一晚上，用C++交不行，结果用C或者G++却过来，哎。。。。这题让我明白，如果用BFS,要保存到达“@”点的步数，bfs一直搜索，直到没发走才停。让后在主函数中进行输出比较。。。

AC代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define N 205char map[N][N];int vis[N][N];int dp1[N][N],dp2[N][N];int x1,y1,x2,y2;struct node{    int x,y,step;}queue[N*N];int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};int n,m;void bfs1(){    struct node now,pre;    queue[0].x=x1;    queue[0].y=y1;    queue[0].step=0;    int e=0,h=1;    while(e<h)    {        pre=queue[e];        for(int i=0;i<4;i++)        {            now.x=pre.x+dir[i][0];            now.y=pre.y+dir[i][1];            now.step=pre.step+1;            if(now.x>=0&&now.x<n&&now.y>=0&&now.y<m&&!vis[now.x][now.y]&&map[now.x][now.y]!='#')            {                dp1[now.x][now.y]=now.step;                vis[now.x][now.y]=1;                queue[h++]=now;            }        }        e++;    }    return;}void bfs2(){    struct node now,pre;    queue[0].x=x2;    queue[0].y=y2;    queue[0].step=0;    int e=0,h=1;    while(e<h)    {        pre=queue[e];        for(int i=0;i<4;i++)        {            now.x=pre.x+dir[i][0];            now.y=pre.y+dir[i][1];            now.step=pre.step+1;            if(now.x>=0&&now.x<n&&now.y>=0&&now.y<m&&!vis[now.x][now.y]&&map[now.x][now.y]!='#')            {                dp2[now.x][now.y]=now.step;                vis[now.x][now.y]=1;                queue[h++]=now;            }        }        e++;    }    return;}int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        memset(dp1,0,sizeof(dp1));        memset(dp2,0,sizeof(dp2));        int i,j;        for(i=0;i<n;i++)        {            scanf("%s",map[i]);        }        for(i=0;i<n;i++)        {            for(j=0;j<m;j++)            {                if(map[i][j]=='M')                {                    x1=i;                    y1=j;                }                if(map[i][j]=='Y')                {                    x2=i;                    y2=j;                }            }        }        memset(vis,0,sizeof(vis));        bfs1();        memset(vis,0,sizeof(vis));        bfs2();        int MIN=N*N;        for(i=0;i<n;i++)        {            for(j=0;j<m;j++)            {                if(map[i][j]=='@'&&dp1[i][j]&&dp2[i][j])                {                    MIN=min(MIN,dp1[i][j]+dp2[i][j]);                }            }        }        printf("%d\n",MIN*11);    }    return 0;}

0 0