DFS【The song of Ice and Fire】

题目描述

    House Lannister of Casterly Rock is one of the Great Houses of Seven Kingdoms, and the principal house of the Westerlands. Their seat is Casterly Rock, though another branch exists that is based in nearby Lannisport. Their sigil is a golden lion on a field of crimson.Their official motto is "Hear Me Roar!" However, their unofficial motto, equally well known, is "A Lannister always pays his debts."The Warden of the West is traditionally a Lannister

    The gold of Casterly Rock,Castamere, the Pendric Hills and the Golden Tooth has made them the wealthiest of the Great Houses, the keep at Casterly Rock sits literally on top of a gold mine, built into the very rock. From that lofty yet uncomfortable position, House Lannister rules over the Westerlands and influences all of Westeros, by virtue of the kingdom’s purse strings.

    One day, Tyrion Lannister come to his gold mine field,which is a large rectangular in the land.If the two gold deposits are adjacent, then it will be one mineral resources. Your tesk is tell him how many mineral resources are contained in this land.

输入格式

The input consists of serveral test cases. For each case contains two positive integer N and M(1 <= M , N <= 100), If N = 0 it means the end of the input; Each of the following N lines contain M characters. Each character is '$' or '#' reflect that there is gold or not.

输出

For each test case, print the number of mineral resources ,and one line one case.

 

样例输入

3 4
##$$
$##$
##$#
1 5
###$$
2 2
##
##
0 0

样例输出

2
1
0

#include<cstring>
#include<iostream>
char x[110][110];
int rem[110][110];
using namespace std;
void dfs(int i,int j)
{
    if(rem[i][j])
        rem[i][j]=0;
    else if(rem[i][j]==0)
        return;
    dfs(i-1,j-1);
    dfs(i+1,j+1);
    dfs(i-1,j+1);
    dfs(i+1,j-1);
    dfs(i-1,j);
    dfs(i+1,j);
    dfs(i,j-1);
    dfs(i,j+1);

}
int main()
{
    int i,j,a,b,sum,k,t;
    while(cin>>a>>b)
    {
        if(a==0&&b==0)
            break;
        memset(rem,0,sizeof(rem));
        sum=0;
        for(i=1; i<=a; i++)
            for(j=1; j<=b; j++)
            {
                cin>>x[i][j];
                if(x[i][j]=='$')
                    rem[i][j]=1;
            }
        for(i=1; i<=a; i++)
            for(j=1; j<=b; j++)
                if(rem[i][j])
                {
                    sum++;
                    dfs(i,j);
                }
        cout<<sum<<endl;
    }
    return 0;
}

（1）：深搜写法，递归  from  魏继鹏

#include<stdio.h>
#include<stdlib.h>
char m[100][100];//数组定义在函数外面，不然你会很悲剧的发现自己的程序无法输入数据。
int a,b;
int  f[8][2]= {{-1,0},{0,1},{1,0},{0,-1},{-1,-1},{-1,1},{1,-1},{1,1}}; //*周围八个点的坐标集合。用两个数组来储存
void dfs(int k,int s)//*调用函数，bfs*//
{
    int i;
    m[k][s]=’#’;//*将搜索过的标记成陆地*//
    for(i=0; i<8; i++)
        if(m[k+f[i][0]][s+f[i][1]]==’$’)//*当周围有金矿的时候，再次搜索其他方位的、*//
            dfs(k+f[i][0],s+f[i][1]);//*不断的调用函数，继续往下一深度搜索，直到把周围的全部搜完*//
}
int main()
{
    int n,i,j,count;
    scanf("%d",&n);
    while(n--)
    {
        count=0;
        scanf("%d%d",&a,&b);
        for(i=1; i<=a; i++)
            for(j=1; j<=b; j++)
                scanf("%c” ,&m[i][j]);
                      for(i=1;i<=a;i++)
                      for(j=1;j<=b;j++)
                  {
                      if(m[i][j]==’$’)
                  {
                      count++;
                      dfs(i,j);//*不断的调用函数*//
                  }
                  }
                      printf("%d\n",count);
                  }
                      return 0;
                  }

AC 代码（2）：广搜写法，队列  from  魏继鹏

#include<stdio.h>
#include<algorithm>
#include<queue>
using namespace std;
int a,b;
char m[100][100];   //地图
int f[8][2]={{-1,0},{0,1},{1,0},{0,-1},{-1,-1},{-1,1},{1,-1},{1,1}}; //八个方向
struct point//*定义队列元素*//
{
int x,y;
};
queue<point> v;//*建立一个队列*//
void bfs(int k,int s)
{
int i;
point L,p,q;
L.x=k;L.y=s;
v.push(L);
m[k][s]=’#’;//*标记*//
while(!v.empty())
{
p=v.front();
for(i=0;i<4;i++)
{
if(m[p.x+f[i][0]][p.y+f[i][1]]==’$’)//*判断是否是金矿或者已经搜索过*//
{
m[p.x+f[i][0]][p.y+f[i][1]]=’$’;
q.x=p.x+f[i][0];q.y=p.y+f[i][1];//*满足条件的放入队列中*//
v.push(q);
}
}
v.pop();//*清空队列*//
}
}
int main()
{
int n,i,j,count;
scanf("%d",&n);
while(n--)
{
count=0;
scanf("%d%d",&a,&b);
for(i=1;i<=a;i++) 
for(j=1;j<=b;j++)
scanf("%c",&m[i][j]);
for(i=1;i<=a;i++)
for(j=1;j<=b;j++)
if(m[i][j]==’$’)
{
count++;
bfs(i,j); //每找到一个符合要求的点，对其进行搜索
}
printf("%d\n",count);
}
return 0;
}