[DFS/BFS]HOJ1797Red and Black

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Red and Black

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 Source : Japan Domestic 2004 Time limit : 1 sec Memory limit : 32 M

Submitted : 834, Accepted : 550

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

解题报告：

此题为典型的搜索题，搜索题有深度优先（DFS）和广度优先（BFS）两种。

DFS递归。很显然，当找到起始状态后，直接向四个方向递归即可。思路很简单代码如下：

#include<stdio.h>#include<string.h>char s[30][30];int maps[30][30],h,w,sum;void dfs(int i,int j){    sum++;    maps[i][j]=1;    if(i+1<w&&maps[i+1][j]==0&&s[i+1][j]=='.')        dfs(i+1,j);    if(j+1<h&&maps[i][j+1]==0&&s[i][j+1]=='.')        dfs(i,j+1);    if(i-1>=0&&maps[i-1][j]==0&&s[i-1][j]=='.')        dfs(i-1,j);    if(j-1>=0&&maps[i][j-1]==0&&s[i][j-1]=='.')        dfs(i,j-1);}int main(){    int i,j;    while(scanf("%d%d",&h,&w)==2&&(h||w)){        memset(maps,0,sizeof(maps));        sum=0;        for(i=0;i<w;i++)            scanf("%s",s[i]);        for(i=0;i<w;i++)            for(j=0;j<h;j++)                if(s[i][j]=='@')                    dfs(i,j);        printf("%d\n",sum);    }    return 0;} 

DFS非递归

因为DFS可以使用栈（stack）实现。所以实现起来就简单了。

#include<iostream>#include<stack>#include<cstdio>using namespace std;struct node{        int x,y;}st;char s[30][30];int maps[30][30];int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};int dfs(node st,int h,int w){    stack<node> q;    int i,cnt=0;    node a,b;    while(!q.empty()){        q.pop();    }    q.push(st);    maps[st.x][st.y]=1;    cnt++;    while(!q.empty()){        a=q.top();        q.pop();        for(i=0;i<4;i++){            b.x=a.x+dir[i][0];            b.y=a.y+dir[i][1];            if(b.x<0||b.y<0||b.x>w-1||b.y>h-1)                continue;            if(maps[b.x][b.y]==0){                q.push(b);                maps[b.x][b.y]=1;                cnt++;            }        }    }    return cnt;}int main(){    int h,w,i,j;    while(scanf("%d%d",&h,&w)==2&&(h||w)){        for(i=0;i<w;i++)            scanf("%s",s[i]);        for(i=0;i<w;i++)            for(j=0;j<h;j++){                if(s[i][j]=='.')                    maps[i][j]=0;                else if(s[i][j]=='#')                    maps[i][j]=1;                else{                    maps[i][j]==0;                    st.x=i;                    st.y=j;            }        }        printf("%d\n",dfs(st,h,w));    }    return 0;} 

BFS

因为BFS可以使用队列（queue）实现。所以代码与DFS非递归相似。

#include<iostream>#include<queue>#include<cstdio>using namespace std;struct node{        int x,y;}st;char s[30][30];int maps[30][30];int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};int bfs(node st,int h,int w){    queue<node> q;    int i,cnt=0;    node a,b;    while(!q.empty()){        q.pop();    }    q.push(st);    maps[st.x][st.y]=1;    cnt++;    while(!q.empty()){        a=q.front();        q.pop();        for(i=0;i<4;i++){            b.x=a.x+dir[i][0];            b.y=a.y+dir[i][1];            if(b.x<0||b.y<0||b.x>w-1||b.y>h-1)                continue;            if(maps[b.x][b.y]==0){                q.push(b);                maps[b.x][b.y]=1;                cnt++;            }        }    }    return cnt;}int main(){    int h,w,i,j;    while(scanf("%d%d",&h,&w)==2&&(h||w)){        for(i=0;i<w;i++)            scanf("%s",s[i]);        for(i=0;i<w;i++)            for(j=0;j<h;j++){                if(s[i][j]=='.')                    maps[i][j]=0;                else if(s[i][j]=='#')                    maps[i][j]=1;                else{                    maps[i][j]==0;                    st.x=i;                    st.y=j;            }        }        printf("%d\n",bfs(st,h,w));    }    return 0;}