poj 2386（深搜或广搜）

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17917		Accepted: 9069

Description

Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors. 

Given a diagram of Farmer John's field, determine how many ponds he has.

Input

* Line 1: Two space-separated integers: N and M 

* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

Output

* Line 1: The number of ponds in Farmer John's field.

Sample Input

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

Sample Output

3

这道题目比较基础，可以深搜也可以广搜

深搜快一点 36ms，广搜47ms。很好的练习题目。

把注释的去掉就是深搜了。

#include <stdio.h>
#include <queue>
using namespace std;
#define N 102

char map[N][N];
int n,m;
struct Node
{
    int x,y;
};
int dis[10][3]={ {0,1},{0,-1},{-1,0},{1,0},{-1,-1},{-1,1},{1,-1},{1,1} };
void search(int a,int b)
{
    /*map[a][b]='.';
    for(int i=0;i<8;i++)
    {
        int tmpx=a+dis[i][0];
        int tmpy=b+dis[i][1];
        if(tmpx>0 && tmpx<=n && tmpy>0 && tmpy<=m && map[tmpx][tmpy]=='W' ){
            //map[tmpx][tmpy]='.';
            search(tmpx,tmpy);
        }
    }*/
    Node tmp,be;
    be.x=a,be.y=b;
    queue<Node> v;
    v.push(be);
    map[a][b]='.';
    while(!v.empty())
    {
        be=v.front();
        v.pop();
        for(int i=0;i<8;i++)
        {
            tmp.x=be.x+dis[i][0];
            tmp.y=be.y+dis[i][1];
            if(tmp.x>0 && tmp.x<=n && tmp.y>0 && tmp.y<=m && map[tmp.x][tmp.y]=='W' )
            {
                v.push(tmp);
                map[tmp.x][tmp.y]='.';
            }
        }
    }
}

int main()
{
    int i,j,count;
    while(~scanf("%d %d",&n,&m))
    {
        getchar();
        count=0;
        for(i=1;i<=n;i++){
            for(j=1;j<=m;j++){
                scanf("%c",&map[i][j]);}  ///缺了括号，害死人啊
            getchar();}
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
                if(map[i][j]=='W')
                {
                    //printf("%d %d\n",i,j);
                    search(i,j);count++;
                }
        printf("%d\n",count);
    }
    return 0;
}

深搜递归写法

#include <stdio.h>
#include <queue>
using namespace std;
#define N 102

char map[N][N];
int n,m;

void search(int i,int j)
{
    if(map[i][j-1]=='W') { map[i][j-1]='.'; search(i,j-1); }
    if(map[i][j+1]=='W') { map[i][j+1]='.'; search(i,j+1); }
    if(map[i-1][j]=='W') { map[i-1][j]='.'; search(i-1,j); }
    if(map[i+1][j]=='W') { map[i+1][j]='.'; search(i+1,j); }
    if(map[i-1][j-1]=='W') { map[i-1][j-1]='.'; search(i-1,j-1); }
    if(map[i-1][j+1]=='W') { map[i-1][j+1]='.'; search(i-1,j+1); }
    if(map[i+1][j-1]=='W') { map[i+1][j-1]='.'; search(i+1,j-1); }
    if(map[i+1][j+1]=='W') { map[i+1][j+1]='.'; search(i+1,j+1); }
}

int main()
{
    int i,j,count;
    while(~scanf("%d %d",&n,&m))
    {
        getchar();
        count=0;
        for(i=1;i<=n;i++){
            for(j=1;j<=m;j++){
                scanf("%c",&map[i][j]);}  ///缺了括号，害死人啊
            getchar();}
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
                if(map[i][j]=='W')
                {
                    //printf("%d %d\n",i,j);
                    search(i,j);count++;
                }
        printf("%d\n",count);
    }
    return 0;
}