HDU 1240 Asteroids!(BFS)

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HDU 1240 Asteroids!(BFS)

http://acm.hdu.edu.cn/showproblem.php?pid=1240

题意:

给你一个n*n*n的三维网格,并且给出初始和终点,要你求最短路径.

分析:

很简单的三维BFS,直接算即可.注意细节.

AC代码:

#include<cstdio>#include<cstring>#include<queue>using namespace std;const int maxn=15;int dr[]={-1,1,0,0,0,0};//前后左右上下int dc[]={0,0,-1,1,0,0};int dh[]={0,0,0,0,-1,1};bool vis[maxn][maxn][maxn];int dist[maxn][maxn][maxn];                        //之前把dist也定义成了bool变量char map[maxn][maxn][maxn];struct Node{    int r,c,h;    Node(int r,int c,int h):r(r),c(c),h(h){}};int n;int r1,c1,h1,r2,c2,h2;int BFS(){    queue<Node> Q;    memset(vis,0,sizeof(vis));    dist[r1][c1][h1]=0;    vis[r1][c1][h1]=true;    if(r1==r2&&c1==c2&&h1==h2) return 0;    Q.push(Node(r1,c1,h1));    while(!Q.empty())    {        Node node=Q.front();Q.pop();        int r=node.r,c=node.c,h=node.h;        for(int d=0;d<6;d++)                    //错误,先前写成了d<4了        {            int nr=r+dr[d],nc=c+dc[d],nh=h+dh[d];            if(nr>=0&&nr<n&&nc>=0&&nc<n&&nh>=0&&nh<n&&!vis[nr][nc][nh]&&map[nh][nr][nc]=='O')            {                vis[nr][nc][nh]=true;                dist[nr][nc][nh]=dist[r][c][h]+1;                Q.push(Node(nr,nc,nh));                if(nr==r2&&nc==c2&&nh==h2) return dist[nr][nc][nh];            }        }    }    return -1;}int main(){    char str[100];    while(scanf("%s%d",str,&n)==2)    {        for(int h=0;h<n;h++)            for(int r=0;r<n;r++)                scanf("%s",map[h][r]);        scanf("%d%d%d%d%d%d",&c1,&r1,&h1,&c2,&r2,&h2);        scanf("%s",str);        int ans=BFS();        if(ans==-1) printf("NO ROUTE\n");        else printf("%d %d\n",n,ans);    }    return 0;}

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