hdu 1241--入门DFS

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11984 Accepted Submission(s): 6952

Problem Description

The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.

Input

The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*', representing the absence of oil, or `@', representing an oil pocket.

Output

For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.

Sample Input

1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5 
****@
*@@*@
*@**@
@@@*@
@@**@
0 0

Sample Output

题目大意：找出连续相连的@有几条由于我是英文渣题目都没看清楚就做错了好久，还理解错了，待会会讲我错哪了；；；；；= 。=

AC代码：

//Oil Deposits
#include <iostream>
#include <stdio.h>
using namespace std;

char map[101][101];
int n, m, sum;

void dfs(int i, int j)
{
    //若该点不可行或越界，返回
    if(map[i][j]!='@' || i<0 || j<0 || i>=m || j>=n) return;
    else    //否则，标记该点为不可行，再往8个方向深搜
    {
        map[i][j] = '*';
        dfs(i-1, j-1);
        dfs(i-1, j);
        dfs(i-1, j+1);
        dfs(i, j-1);
        dfs(i, j+1);
        dfs(i+1, j-1);
        dfs(i+1, j);
        dfs(i+1, j+1);
    }
}

int main()
{
    int i, j;
    while(cin>>m>>n)
    {
        if(m==0 || n==0) break;
        sum = 0;
        for(i = 0; i < m; i++)
            for(j = 0; j < n; j++)
                cin>>map[i][j];
        for(i = 0; i < m; i++)
        {
            for(j = 0; j < n; j++)
            {
                if(map[i][j] == '@')
                {
                    dfs(i, j);
                    sum++;
                }
            }
        }
        cout<<sum<<endl;
    }

    return 0;
}

我之前的做法（错的）：

错误：一开始没注意到对角线也可以后来修正了，但是还是WA

错误2：WA的原因就是我多此一举还用hash存我看错题目以为相同长度的只能算一次（T_T只看样例去理解题目意思的好桑不起啊~~）

//============================================================================
// Name        : 2333.cpp
// Author      : zjx
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
using namespace std;
char op[105][105];
bool num[105][105];
bool hash[10010];
int dfs(int n,int m){
	if(num[n][m]||op[n][m]=='*'){
	return 0;
	}
	num[n][m]=true;
	int sum;
	sum=1+dfs(n-1,m)+dfs(n+1,m)+dfs(n,m+1)+dfs(n,m-1)+dfs(n+1,m+1)+dfs(n-1,m-1)+dfs(n+1,m-1)+dfs(n-1,m+1);
	return sum;
}
int main()
{
	int a,b,i,j;
	while(cin>>a>>b){
		if(a==0&&b==0)break;
		memset(num,false,sizeof(num));
		memset(hash,false,sizeof(hash));
	for(i=1;i<=a;i++)
	for(j=1;j<=b;j++){
		cin>>op[i][j];
	}
	int sum=0;int temp;
	for(i=0;i<=a+1;i++){op[i][0]='*';op[i][b+1]='*';}
	for(j=0;j<=b+1;j++){op[0][j]='*';op[a+1][j]='*';}
	for(i=1;i<=a;i++)
		for(j=1;j<=b;j++){
			temp=dfs(i,j);
			if(temp!=0&&hash[temp]==false){
				sum++;hash[temp]=true;
			}
		}
	cout<<sum<<endl;
}
	return 0;
}

today lazy . tomorrow die .