HDU 1242 rescue

It should be known to the Chinese.

Train of Thought: This question is quite easy and wrong without careful consideration. Like me, I didn't dare to use the priority queue after I finished learning STL, so I still suffered a loss!

That is, you need to find a from R.

If you simply use a queue

3 3

R ..

# X.

# When a is used, it is easy to make a mistake, and may output 5 or 4. At this time, it is wrong, so we can only use the priority queue for optimization, so that the number of steps is less than the priority queue, if you are not familiar with it, you can check STL.

Code:

#include<cstdio>#include<cstring>#include<queue>using namespace std;char str[205][205];int vis[205][205];int dir[4][2]={{-1,0},{0,-1},{1,0},{0,1}};int n,m;struct node{int x,y;int step;friend bool operator<(const node a,const node b){return a.step>b.step;}};int bfs(int x,int y){int i;node st,ed;priority_queue<node>q;memset(vis,0,sizeof(vis));    st.x=x;st.y=y;st.step=0;vis[st.x][st.y]=1;q.push(st);while(!q.empty()){st=q.top();q.pop();if(str[st.x][st.y]=='r')            return st.step;        for(i=0;i<4;i++){ed.x=st.x+dir[i][0];ed.y=st.y+dir[i][1];if((0<=ed.x&&ed.x<n&&0<=ed.y&&ed.y<m&&str[ed.x][ed.y]!='#')&&!vis[ed.x][ed.y]){vis[ed.x][ed.y]=1;if(str[ed.x][ed.y]=='x')                    ed.step=st.step+2;else                    ed.step=st.step+1;q.push(ed);}}}return -1;}int main(){int i,j;int x,y,sum;while(scanf("%d%d",&n,&m)!=EOF){for(i=0;i<n;i++)            scanf("%s",str[i]);for(i=0;i<n;i++)            for(j=0;j<m;j++)                if(str[i][j]=='a'){x=i;y=j;break;}sum=bfs(x,y);if(sum==-1)            printf("Poor ANGEL has to stay in the prison all his life.\n");else            printf("%d\n",sum);}return 0;}

I saw another code from the Internet, which can be used for reference:

#include<iostream>#include<cstring>#include<queue>using namespace std;#define INF 99999char maze[205][205];long len[205][205];int mov[4][2]={{1,0},{-1,0},{0,1},{0,-1}};void BFS(int xa,int ya){    int x0,y0,j,p,q;    queue<int> x;    queue<int> y;    x.push(xa);    y.push(ya);    while(!x.empty()&&!y.empty())    {        x0=x.front();y0=y.front();        x.pop();y.pop();        for(j=0;j<4;j++)        {            p=x0+mov[j][0];q=y0+mov[j][1];            if(maze[p][q]=='.'||maze[p][q]=='r'||maze[p][q]=='x')            {                if(maze[p][q]=='x')                {                    if(len[x0][y0]+2<len[p][q]||len[p][q]==0)                    {                        len[p][q]=len[x0][y0]+2;                        x.push(p);y.push(q);                    }                }                else if(len[x0][y0]+1<len[p][q]||len[p][q]==0)                {                    len[p][q]=len[x0][y0]+1;                    x.push(p);y.push(q);                }            }        }    }}int main(){    int m,n,i,j,a,b;    long ans;    while(cin>>m>>n)    {        ans=INF;        memset(maze,'*',sizeof(maze));        memset(len,0,sizeof(len));        for(i=1;i<=m;i++)            for(j=1;j<=n;j++)            {                cin>>maze[i][j];                if(maze[i][j]=='a')                {                    a=i;b=j;                }            }        BFS(a,b);        for(i=1;i<=m;i++)            for(j=1;j<=n;j++)                if(maze[i][j]=='r'&&ans>len[i][j]&&len[i][j])                    ans=len[i][j];        if(ans==INF)            cout<<"Poor ANGEL has to stay in the prison all his life."<<endl;        else            cout<<ans<<endl;    }    return 0;}

HDU 1242 rescue