csdn1780 优先队列BFS

题目链接：点击打开链接

第一次是求正常的BFS

第二次是每次要转弯的BFS

求第二个只要记住每个点最多是由相邻的个点走过来的就要好了，用vis2标记四个方向，每个方向最多走一次

#include <iostream>#include<cstdio>#include<cstring>#include<cstdio>#include<cstring>#include<queue>#include<cstdlib>using namespace std;const int maxn=500+10;const int xx[]={0,0,1,-1};const int yy[]={1,-1,0,0};int m[maxn][maxn],vis1[maxn][maxn],vis2[maxn][maxn][4];int N,M,c1,c2,r1,r2; struct node{    int x,y,dis,dir;//dir表示方向       friend bool operator <(node a,node b)    {       return a.dis>b.dis;    }};int check(int x,int y){   if(x>0&&x<=N&&y>0&&y<=M&&m[x][y]!=-1)return 1;   return 0;}int bfs1(){    if(c1 == c2&& r1 == r2)        return 0;    priority_queue<node> q;    memset(vis1,0,sizeof(vis1));    node temp,next;    temp.x = r1 ;    temp.y = c1 ;    temp.dis = m[r1][c1];    q.push(temp);    vis1[r1][c1] = 1;    while(!q.empty())    {        temp = q.top();        q.pop();        if(temp.x == r2&&temp.y == c2)            return temp.dis;        for(int i = 0 ; i < 4; i ++)        {            int nx = temp.x + xx[i];            int ny = temp.y + yy[i];            next.x=nx,next.y=ny;            if(check(nx,ny)&&!vis1[nx][ny])            {                vis1[nx][ny] = 1;               next.dis = temp.dis +m[nx][ny];                q.push(next);            }        }    }    return -1;}int bfs2(){    if(c1 == c2&& r1 == r2)        return 0;    priority_queue<node> q;    memset(vis2,0,sizeof(vis2));    node temp,next;    temp.x = r1 ;    temp.y = c1 ;    temp.dis = m[r1][c1];    temp.dir=-1;    q.push(temp);    vis1[r1][c1] = 1;    while(!q.empty())    {        temp = q.top();        q.pop();        if(temp.x == r2&&temp.y == c2)            return temp.dis;        for(int i = 0 ; i < 4; i ++)        {            int nx = temp.x + xx[i];            int ny = temp.y + yy[i];            next.x=nx,next.y=ny;            if(check(nx,ny)&&!vis2[nx][ny][i]&&temp.dir!=i)            {                vis2[nx][ny][i] = 1;                next.dir=i;               next.dis = temp.dis +m[nx][ny];                q.push(next);            }        }    }    return -1;}int main(){ int ans=0,i,j; char s[100]; while(~scanf("%d %d %d %d %d %d",&N,&M,&r1,&c1,&r2,&c2)) {     memset(m,0x3f,sizeof(m));    for(i=1;i<=N;i++)    for(j=1;j<=M;j++)    {        scanf("%s",s);        if(s[0]=='*')m[i][j]=-1;        else m[i][j]=atoi(s);    }     printf("Case %d: %d %d\n",++ans,bfs1(),bfs2()); }    return 0;}