hdu 1312 Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6064    Accepted Submission(s): 3854

Problem Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above. 

 

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile 
'#' - a red tile 
'@' - a man on a black tile(appears exactly once in a data set) 

 

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself). 

 

Sample Input

6 9....#......#..............................#@...#.#..#.11 9.#..........#.#######..#.#.....#..#.#.###.#..#.#..@#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###...@...###.###..#.#....#.#..0 0

 

Sample Output

4559613

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

Recommend

Eddy

解题思路：第一次完全靠自己写出来的搜索题，纪念一下，用的是BFS，用了模版类"优先队列"，把每个搜索到的数据放入队列中，然后计数器的值增加一，把搜索过的值用vis数组进行标记，这样可以不改变mp数组的值，避免常见的错误发生。

还有当初wa的原因时像

@#..

#...

....

....

答案为1，而自己的答案却是0，希望对wa的同学有所帮助。

代码：

#include<cstdio>#include<queue>#include<cstring>using namespace std;struct node{int x,y;int num;friend bool operator < (node n1,node n2){return n2.x<n1.x;}};int mp[25][25],n,m,si,sj,ans;int dir[4][2]={1,0,-1,0,0,1,0,-1};int vis[25][25];int notinmap(int a,int b){if(a<0||a>=m||b<0||b>=n)return 1;if(mp[a][b]=='#')return 1;return 0;}void bfs(){priority_queue<node>Q;while(!Q.empty()) Q.pop();memset(vis,0,sizeof(vis));node cur,next;cur.x=si;cur.y=sj;cur.num=0;Q.push(cur);while(!Q.empty()){cur=Q.top();Q.pop();for(int i=0;i<4;i++){next.x=cur.x+dir[i][0];next.y=cur.y+dir[i][1];if(notinmap(next.x,next.y)) continue;if(vis[next.x][next.y]) continue; Q.push(next);vis[next.x][next.y]=1;ans++;}}}int main(){char str[25];while(scanf("%d%d",&n,&m)&&(n+m)){getchar();ans=0,si=0,sj=0;for(int i=0;i<m;i++){scanf("%s",str);for(int j=0;str[j];j++){if(str[j]=='@'){si=i;sj=j;}mp[i][j]=str[j];}}//printf("%d %d\n",si,sj);bfs();if(ans==0)printf("%d\n",1);elseprintf("%d\n",ans);}return 0;}