Depth-first search DFS
The so-called depth-first search, popular point is to understand the head went one way - do not hit the brick wall does not look back.
Let's look at a full array issue, now on 123 full-arranged, now holding the hands of small hum 123 three cards, he will give this three cards into three boxes, each filled It is not a fully arranged yet?
But in the end every time the card is first put 1 or 3 card it?
Well think small, I put it in order, and each time put the card in the order 1.2.3. So he went to the former No. 1 box 1 into the card, went to No. 2 before the card 2 into the box, went to No. 3 before the card 3 into the box, but little came fourth box ... hum card has been done and
La. Then arose a full array " 123 " friends.
But there is a good variety of circumstances ah, so have the full array of small hum latter to return immediately, before he returned to the No. 3 box, retrieve the card 3, see also not put other cards, revealing a small hum hands now In addition to the card 3 is not no other external cards, so little hum continue
Go back to the No. 2 box before, 2 small hum recover the card, which is the hand of the little hum has two cards, so he put the card 3 No. 2 box, then put away to take a step back Three box, put the card 2, the top four went to the box, of course not
Fourth box, so he produces a full array " 132 "
Thus according to the simulation, it will in turn generate a full-arrangement "213" "231" "312" "321"
Said a long time, we look at how to use code to achieve it
I number of cards in the first step of the boxes
for(i = 1; i <=n; i++)
{
a[step] = i; //将卡片i放入第step个盒子里
}But there's a problem, how a card has been placed in a box in the other, you can not put the current box. So, we need a book [] array to mark this card hand is there
for(i = 1; i <= n; i++)
{
if(book[i] == 0) / /如果手上有这张卡片
{
a[step] = i; //将卡片i放入第step个盒子里
book[i] = 1; //此时手上已经没有这张卡了 标记一下
}
}After processing method of the step boxes we want to go back one step, the first processing step + 1 boxes, apparently the first processing step + 1 boxes and boxes as the first step, then we will just deal with the first step of boxes written as a function method
void dfs(int step)
{
for(i = 1; i <= n; i++)
{
if(book[i] == 0) / /如果手上有这张卡片
{
a[step] = i; //将卡片i放入第step个盒子里
book[i] = 1; //此时手上已经没有这张卡了 标记一下
}
}
}Treatment step + 1 boxes only dfs (step + 1) on the line
void dfs(int step)
{
for(i = 1; i <= n; i++)
{
if(book[i] == 0) / /如果手上有这张卡片
{
a[step] = i; //将卡片i放入第step个盒子里
book[i] = 1; //此时手上已经没有这张卡了 标记一下
}
dfs(step+1);//处理下个盒子
book[i] = 0; //这步很重要！！往回走时要将之前的卡片收回！
}
}book[i] = 0;是将之前的卡片收回，如果不把刚才放入盒子的卡片收回，那么就无法再进行下一次摆放。
还剩最后一个问题，什么时候输出一个满足条件的序列呢？其实当我们走到第n+1个盒子前时，说明前n个盒子都已经放好卡片了，这时候我们就要return返回了
void dfs(int step)
{
if (step == n + 1) //判断边界
{
for (i = 1; i <= n; i++)
cout << a[i] << " ";
cout << endl;
return;//返回上一个盒子前
}
else
{
for (i = 1; i <= n; i++)
{
if (book[i] == 0) / /如果手上有这张卡片
{
a[step] = i; //将卡片i放入第step个盒子里
book[i] = 1; //此时手上已经没有这张卡了 标记一下
}
dfs(step + 1);//处理下个盒子
book[i] = 0; //这步很重要！！往回走时要将之前的卡片收回！
}
}
}讲到这里想必大家已经对深度优先搜索有一定了解了吧，深搜使用就是递归和回溯的思想。
广度优先搜索BFS
讲完了dfs，我们再来看看什么是广度优先搜索。
所谓广度优先搜索就是一种层层递进的搜索。
我们来看一张图。这是一个迷宫，我们用一个二维数据储存它。
现在小哼站在（1 1）处，它可以向下或向右走，如何才能到达终点呢?已经学会深搜的你因该很快就想到办法了，但这里我们用另一种方法BFS，“一层层”扩展找到终点。扩展时每发现一个点就将这个点放入队列中，知道走到终点位置。和深搜一样，我们也需要一个book数组记录是否走过这个点。
我们每对一个点扩展完毕，这个点就没用了，所以要将这个点出队，对下个点扩展。
理解了这点后代码就很容易实现了^_^。
完整代码
#include<bits/stdc++.h>
using namespace std;
struct node
{
int x, y;
int s;//步数
int f;//父亲再队列中的编号
};
int main()
{
queue<node> que;
int a[51][51] = { 0 };
int book[51][51] = { 0 };
int next[4][2] =
{
{0,1}, //向右走
{1,0},//向下走
{0,-1},//向左走
{-1,0}//向上走
};
int n, m, i, j, k, flag,p,q;
node start, tnode;
//输入迷宫的行列 起始点
cin >> n >> m >> start.x >> start.y;
//输入迷宫图
for (i = 1; i <= n; i++)
for (j = 1; j <= m; j++)
cin >> a[i][j];
cin >> p >> q;//输入终点
book[start.x][start.y] = 1;//将起始点标记为走过
//如果队列不为空
while (!que.empty())
{
//枚举四个方向
for (int k = 0; k <= 3; k++)
{
//计算下一步坐标
tnode.x = que.front().x + next[k][0];
tnode.y = que.front().y + next[k][1];
//判断是否出界
if (tnode.x<1 || tnode.x>n || tnode.y<1 || tnode.y>m)
continue;
//判断是否是陆地或已经走过
if (mapp[tnode.x][tnode.y] > 0 && book[tnode.x][tnode.y] == 0)
{
sum++;
//每个点只入队一次，标记走过
book[tnode.x][tnode.y] = 1;
int ts = que.back().s;//记录父亲的步数
//将该点入队
que.push(tnode);
que.back().s = ts+1; //步数是父亲步数加1
}
if (tnode.x == p && tnode.y == q)
{
flag = 1;
break;
}
}
if (flag == 1)
break;
que.pop();//队首出队
}
cout << que.back().s << endl;
system("pause");
return 0;
}练习
学完DFS和BFS后来练习一下吧。
题目：宝岛探索
小哼通过一种秘密方法得到一张不完整的钓鱼岛航拍图。小哼决定去钓鱼岛探险。
下面的10*10的矩阵就是该航拍图。图中数字表示海拔，0表示海洋，1-9都表示陆地。
小哼的飞机会降落再（6，8），现在需要计算小哼降落岛的面积。此处我们把降落点上
下左右相邻岛视为同一岛屿。
输入
输出
BFS代码如下：
#include<bits/stdc++.h>
using namespace std;
struct node
{
int x;
int y;
};
int main()
{
queue<node> que;
int mapp[51][51];
int book[51][51];
int i, j, n, m,sum;
node start, tnode;
memset(book, 0, sizeof(book));
memset(mapp, 0, sizeof(mapp));
//输入地图的行列，初始坐标
cin >> n >> m >> start.x >> start.y;
//输入地图
for(i=1;i<=n;i++)
for (j = 1; j <= m; j++)
{
cin >> mapp[i][j];
}
int next[4][2]=
{
{0,1}, //向右走
{1,0},//向下走
{0,-1},//向左走
{-1,0}//向上走
};
//向队列插入降落的起始坐标
que.push(start);
sum = 1;
book[start.x][start.y] = 1;
//当队列不为空时循环
while (!que.empty())
{
//枚举四个方向
for (int k = 0; k <= 3; k++)
{
//计算下一步坐标
tnode.x = que.front().x + next[k][0];
tnode.y= que.front().y + next[k][1];
//判断是否出界
if (tnode.x<1 || tnode.x>n || tnode.y<1 || tnode.y>m)
continue;
//判断是否是陆地或已经走过
if (mapp[tnode.x][tnode.y] > 0 && book[tnode.x][tnode.y] == 0)
{
sum++;
//每个点只入队一次，标记走过
book[tnode.x][tnode.y] = 1;
//将该点入队
que.push(tnode);
}
}
//队首出队，下个点继续搜素
que.pop();
}
cout << sum << endl;
getchar();
getchar();
return 0;
}DFS代码如下：
#include<bits/stdc++.h>
using namespace std;
struct node
{
int x;
int y;
};
int mapp[51][51];
int book[51][51];
int n, m,sum;
void dfs(int x,int y)
{
int next[4][2] =
{
{0,1}, //向右走
{1,0},//向下走
{0,-1},//向左走
{-1,0}//向上走
};
int tx, ty, k;
//枚举四个方向
for (k = 0; k <= 3; k++)
{
tx = x + next[k][0];
ty = y + next[k][1];
//判断是否出界
if (tx<1 || tx>n || ty<1|| ty>m)
continue;
//判断是否是陆地或已经走过
if (mapp[tx][ty] > 0 && book[tx][ty] == 0)
{
sum++;
//标记走过
book[tx][ty] = 1;
dfs(tx, ty); //执行下一步
}
}
return;
}
int main()
{
int i, j,sx,sy;
memset(book, 0, sizeof(book));
memset(mapp, 0, sizeof(mapp));
cin >> n >> m >> sx >> sy;
//输入地图
for(i=1;i<=n;i++)
for (j = 1; j <= m; j++)
{
cin >> mapp[i][j];
}
book[sx][sy] = 1;
sum = 1;
dfs(sx, sy);
cout << sum << endl;
getchar();
getchar();
return 0;
}拓展：如何求有多少个独立岛屿呢？
即求独立子图的个数，这就是Floodfill漫水填充法，又叫种子填充法。Windows下“画图”中的油漆桶工具就是基于这个算法的。
我们将独立岛屿进行染色，所以增加参数color。
代码如下
#include<bits/stdc++.h>
using namespace std;
struct node
{
int x;
int y;
};
int mapp[51][51];
int book[51][51];
int n, m,sum;
void dfs(int x,int y,int color)
{
mapp[x][y] = color;//表示小哼来过这个岛
int next[4][2] =
{
{0,1}, //向右走
{1,0},//向下走
{0,-1},//向左走
{-1,0}//向上走
};
int tx, ty, k;
//枚举四个方向
for (k = 0; k <= 3; k++)
{
tx = x + next[k][0];
ty = y + next[k][1];
//判断是否出界
if (tx<1 || tx>n || ty<1|| ty>m)
continue;
//判断是否是陆地或已经走过
if (mapp[tx][ty] > 0 && book[tx][ty] == 0)
{
sum++;
//标记走过
book[tx][ty] = 1;
dfs(tx, ty,color); //执行下一步
}
}
return;
}
int main()
{
int i, j,sx,sy,color=0;
memset(book, 0, sizeof(book));
memset(mapp, 0, sizeof(mapp));
cin >> n >> m ;
//输入地图
for(i=1;i<=n;i++)
for (j = 1; j <= m; j++)
{
cin >> mapp[i][j];
}
//对每个大于0（未被染色）的点尝试dfs染色
for (i = 1; i <= n; i++)
{
for (j = 1; j <= m; j++)
{
if (mapp[i][j] > 0)//如果没被染色
{
color--;//color减一表示一种新的颜色
book[i][j] = 1; //标记走过
dfs(i, j, color);
}
}
}
for (i = 1; i <= n; i++)
{
for (j = 1; j <= m; j++)
{
cout<<mapp[i][j]<<" ";
}
cout << endl;
}
cout <<"共有独立岛屿："<< -color<<" 个" << endl;
getchar();
getchar();
return 0;
}