Oh, my goddess

Oh, my goddess时间限制：3000 ms  |  内存限制：65535 KB难度：3描述Shining Knight is the embodiment of justice and he has a very sharp sword can even cleavewall. Many bad guys are dead on his sword.One day, two evil sorcerer cgangee and Jackchess decided to give him some colorto see. So they kidnapped Shining Knight's beloved girl--Miss Ice! They built a M x Nmaze with magic and shut her up in it.Shining Knight arrives at the maze entrance immediately. He can reach any adjacent emptysquare of four directions -- up, down, left, and right in 1 second. Or cleave one adjacent wall in 3seconds, namely,turn it into empty square. It's the time to save his goddess! Notice: ShiningKnight won't leave the maze before he find Miss Ice.输入The input consists of blocks of lines. There is a blank line between two blocks.The first line of each block contains two positive integers M <= 50 and N <= 50separated by one space. In each of the next M lines there is a string of length N contentsO and #.O represents empty squares. # means a wall.At last, the location of Miss Ice, ( x, y ). 1 <= x <= M, 1 <= y <= N.(Shining Knight always starts at coordinate ( 1, 1 ). Both Shining and Ice's locationguarantee not to be a wall.)输出The least amount of time Shining Knight takes to save hisgoddess in one line.样例输入3 5O##########O#O#3 4样例输出14


个人理解：该题使用优先队列处理结果时间运行语言Accepted40324C/C++
#include<stdio.h>#include<queue>#include<string.h>using namespace std;void bb();int x,y,ex,ey,m,n;char vc[51][51];int cc[51][51],vx[4]={-1,0,0,1},vy[4]={0,-1,1,0};struct dear{    int x,y,time;     friend bool operator < (dear a,dear b)//由小到大排列    {        return a.time>b.time;    }};int main(){    int i,j;    while(~scanf("%d%d",&m,&n))  {    for(i=1;i<=m;i++)    {        getchar();//回收回车键储存空间        for(j=1;j<=n;j++)        {            scanf("%c",&vc[i][j]);        }    }    scanf("%d%d",&ex,&ey);    bb();  }    return 0;}void bb(){    int i;    memset(cc,0,sizeof(cc));    priority_queue<dear> pq;    dear st={1,1,0};    pq.push(st);    cc[1][1]=1;    while(pq.empty()!=true)    {        st=pq.top();        pq.pop();        if(st.x==ex&&st.y==ey)        {            printf("%d\n",st.time);            return;        }        for(i=0;i<4;i++)        {            int cx,cy;            cx=st.x+vx[i];            cy=st.y+vy[i];            if(cx<1||cx>m||cy<1||cy>n)            {                continue;            }            if(!cc[cx][cy])            {                int ct=st.time+1;                if(vc[cx][cy]=='#')                    ct+=3;                    dear cp={cx,cy,ct};                    cc[cx][cy]=1;                    pq.push(cp);            }        }    }}