LightOJ-1066-bfs

来源：互联网发布：linux 输出文件夹大小编辑：程序博客网时间：2024/06/06 00:59

题目大意：一张坐标图上有若干个食物，每个都用一个英文字母代表，现在需要按字典序一次寻找每个食物问最少需要多少路程；

题目解析：bfs即可，依次寻找，每次寻找成功之后就把上一个置位‘.’，表示可以通过；

AC代码：

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<queue>using namespace std;int graph[11][11];bool vis[11][11];int dx[4]={-1,0,0,1};int dy[4]={0,1,-1,0};struct code{int x,y,num;code(int a,int b,int c){x=a;y=b;num=c;}};queue<code>q;int  main(){int cas,c,i,j;char s1[17];scanf("%d",&cas);for(c=1;c<=cas;c++){while(!q.empty())q.pop();int n,cnt=0;scanf("%d",&n);for(i=1;i<=n;i++){scanf("%s",&s1[1]);for(j=1;j<=n;j++){if(s1[j]=='.')graph[i][j]=0;else if(s1[j]=='#')graph[i][j]=-1;else {graph[i][j]=s1[j]-'A'+1;cnt++;if(graph[i][j]==1){q.push(code(i,j,0));}}}}int flag=1,tar=2,ans=0;for(i=1;i<=cnt-1;i++){memset(vis,0,sizeof(vis));int temp1,temp2;temp1=q.front().x;temp2=q.front().y;vis[temp1][temp2]=1;int f=0;int tar_x,tar_y;while(!q.empty()){code pos=q.front();q.pop();for(j=0;j<4;j++){int xx=pos.x+dx[j];int yy=pos.y+dy[j];//cout<<xx<<" "<<yy<<endl;if(xx<1||xx>n||yy<1||yy>n)continue;if(graph[xx][yy]==tar&&!vis[xx][yy]){ans+=pos.num+1;f=1;tar_x=xx;tar_y=yy;break;}else if(graph[xx][yy]==0&&!vis[xx][yy]){q.push(code(xx,yy,pos.num+1));vis[xx][yy]=1;}}if(f==1)break;}if(f==1){while(!q.empty())q.pop();tar++;q.push(code(tar_x,tar_y,0));graph[temp1][temp2]=0;}else {flag=0;break;}}if(flag)printf("Case %d: %d\n",c,ans);else printf("Case %d: Impossible\n",c);}return 0;}

0 0