一、深度优先搜索DFS
深度优先搜索DFS的关键思想是：当下应该怎么做（每个方法都试一遍），这一步解决后，进入下一步，下一步的解决方法和这一步的解决方法是一样的
DFS的基本模型
void dfs(int step)
{
判断边界
尝试每一种可能 for(i=1;i<=n;i++)
{
继续下一步 dfs(step+1)
}
返回
}
1.1全排列
1 //输入一个数n
2 //输出1-n的全排列
3 #include <stdio.h>
4 int n, book[10], a[10];
5 void dfs(int step)
6 {
7 int i;
8 if (step > n)
9 {
10 for (i = 1;i <= n;i++)
11 {
12 printf("%d ",a[i]);
13 }
14 printf("\n");
15 return;
16 }
17 for (i = 1;i <= n;i++)
18 {
19 if (book[i] == 0)
20 {
21 a[step] = i;
22 book[i] = 1;
23 dfs(step+1);
24 book[i] = 0;
25 }
26 }
27 return;
28 }
29 int main()
30 {
31 scanf("%d",&n);
32 dfs(1);
33 return 0;
34 }
1.2两个三位数相加的结果是三位数，这9个数字各不相同，_ _ _ + _ _ _ = _ _ _
1 #include <stdio.h>
2 int total = 0;
3 int book[10], a[10];
4 void dfs(int step)
5 {
6 int i;
7 if (step == 10)
8 {
9 if (100 * (a[1] + a[4]) + 10 * (a[2] + a[5]) + a[3] + a[6] == 100 * a[7] + 10 * a[8] + a[9])
10 {
11 total++;
12 printf("%d%d%d + %d%d%d = %d%d%d\n",a[1],a[2],a[3],a[4],a[5],a[6],a[7],a[8],a[9]);
13 }
14 return;
15 }
16 for (i = 1;i < 10;i++)
17 {
18 if (book[i] == 0)
19 {
20 book[i] = 1;
21 a[step] = i;
22 dfs(step+1);
23 book[i] = 0;
24 }
25 }
26 return;
27 }
28 int main()
29 {
30 dfs(1);
31 printf("%d\n",total/2);
32 return 0;
33 }
1.3走迷宫，解救小哈
1 #include <stdio.h>
2 int n, m, map[10][10], book[10][10],endx, endy,minstep=978978;
3 void dfs(int x, int y, int step)
4 {
5 int i,tx,ty;
6 int next[4][2] = { {0,1},{1,0},{0,-1},{-1,0} };
7 if (x == endx&&y == endy)
8 {
9 if (minstep > step)
10 minstep = step;
11 return;
12 }
13 for (i = 0;i < 4;i++)
14 {
15 tx = x + next[i][0];
16 ty = y + next[i][1];
17 if (tx<1 || ty<1 || tx>n || ty>m)
18 continue;
19 if (map[tx][ty] == 0 && book[tx][ty] == 0)
20 {
21 book[tx][ty] = 1;
22 dfs(tx,ty,step+1);
23 book[tx][ty] = 0;
24 }
25 }
26 return;
27 }
28 int main()
29 {
30 int i, j, startx, starty;
31 scanf("%d%d",&n,&m);
32 for (i = 1;i <= n;i++)
33 {
34 for (j = 1;j <= m;j++)
35 {
36 scanf("%d",&map[i][j]);
37 }
38 }
39 scanf("%d%d%d%d",&startx,&starty,&endx,&endy);
40
41 dfs(startx,starty,0);
42
43 printf("%d\n",minstep);
44 return 0;
45 }
二、广度优先搜索BFS
深度优先搜索DFS——递归
广度优先搜索BFS——队列
继续走迷宫
1 #include <stdio.h>
2 int n, m, endx, endy;
3 int map[10][10], book[10][10];
4 struct node
5 {
6 int x;
7 int y;
8 int step;
9 };
10 void bfs(int startx,int starty)
11 {
12 struct node queue[100];
13 int head, tail;
14 int next[4][2] = { {0,1},{1,0},{0,-1},{-1,0} };
15 int i, tx, ty, flag = 0;
16 head = 1;tail = 1;
17 queue[tail].x = startx;
18 queue[tail].y = starty;
19 queue[tail].step = 0;
20 tail++;
21 book[startx][starty] = 1;
22 while (head<tail)
23 {
24 for (i = 0;i < 4;i++)
25 {
26 tx = queue[head].x + next[i][0];
27 ty = queue[head].y + next[i][1];
28 if (tx<1 || ty<1 || tx>n || ty>m)
29 {
30 continue;
31 }
32 if (map[tx][ty] == 0 && book[tx][ty] == 0)
33 {
34 book[tx][ty] = 1;
35 queue[tail].x = tx;
36 queue[tail].y = ty;
37 queue[tail].step = queue[head].step + 1;
38 tail++;
39 }
40 if (tx == endx&&ty == endy)
41 {
42 flag = 1;
43 break;
44 }
45 }
46 if (flag == 1)
47 break;
48 head++;
49 }
50 printf("%d\n",queue[tail-1].step);
51
52 }
53 int main()
54 {
55 int i, j,startx,starty;
56 scanf("%d%d",&n,&m);
57 for (i = 1;i <= n;i++)
58 {
59 for (j = 1;j <= m;j++)
60 {
61 scanf("%d",&map[i][j]);
62 }
63 }
64 scanf("%d%d%d%d",&startx,&starty,&endx,&endy);
65 bfs(startx,starty);
66 return 0;
67 }
三、再解炸弹人
现在炸弹不是想放在那里就能放在那里的了，必须由小人能够走到的地方才能放置炸弹。比如下面这个例子小人默认站在(3,3)这个位置。请问放在何处最多可以消灭多个敌人。
输入
13 13 3 3
#############
#GG.GGG#GGG.#
###.#G#G#G#G#
#.......#..G#
#G#.###.#G#G#
#GG.GGG.#.GG#
#G#.#G#.#.#.#
##G...G.....#
#G#.#G###.#G#
#...G#GGG.GG#
#G#.#G#G#.#G#
#GG.GGG#G.GG#
#############
输出
将炸弹放置在(7,11)处，最多可以消灭10个敌人。
1 #include <stdio.h>
2 char map[20][20];
3 int n, m, book[20][20];
4 int max=0, maxx, maxy;
5 struct node
6 {
7 int x;
8 int y;
9 };
10 int fun(int x,int y)
11 {
12 int sum = 0, k;
13 k = x;
14 while (map[k][y] != ‘#‘&&k >= 0)
15 {
16 if (map[k][y] == ‘G‘)
17 sum++;
18 k--;
19 }
20 k = x;
21 while (map[k][y] != ‘#‘&&k < n)
22 {
23 if (map[k][y] == ‘G‘)
24 sum++;
25 k++;
26 }
27 k = y;
28 while (map[x][k] != ‘#‘&&k >= 0)
29 {
30 if (map[x][k] == ‘G‘)
31 sum++;
32 k--;
33 }
34 k = y;
35 while (map[x][k] != ‘#‘&&k <m)
36 {
37 if (map[x][k] == ‘G‘)
38 sum++;
39 k++;
40 }
41 return sum;
42 }
43 void dfs(int x,int y)
44 {
45 int next[4][2] = { {0,1},{1,0},{0,-1},{-1,0} };
46 int i,tx,ty;
47 int sum = fun(x, y);
48 if (max < sum)
49 {
50 max = sum;
51 maxx = x;
52 maxy = y;
53 }
54 for (i = 0;i < 4;i++)
55 {
56 tx = x + next[i][0];
57 ty = y + next[i][1];
58 if (tx < 0 || tx >= n || ty < 0 || ty >= m)
59 continue;
60 if (map[tx][ty] == ‘.‘&&book[tx][ty] == 0)
61 {
62 book[tx][ty] = 1;
63 dfs(tx,ty);
64 }
65 }
66 return;
67 }
68 void bfs(int x,int y)
69 {
70 struct node queue[200];
71 int head, tail,i,tx,ty,sum;
72 int next[4][2] = { {0,1},{1,0},{0,-1},{-1,0} };
73 head = 0;
74 tail = 0;
75 queue[tail].x = x;
76 queue[tail].y = y;
77 tail++;
78 while (head < tail)
79 {
80 for (i = 0;i < 4;i++)
81 {
82 tx = queue[head].x + next[i][0];
83 ty = queue[head].y + next[i][1];
84 if (tx < 0 || tx >= n || ty < 0 || ty >= m)
85 continue;
86 if (map[tx][ty] == ‘.‘&&book[tx][ty] == 0)
87 {
88 queue[tail].x = tx;
89 queue[tail].y = ty;
90 tail++;
91 sum = fun(tx, ty);
92 book[tx][ty] = 1;
93 if (sum > max)
94 {
95 max = sum;
96 maxx = tx;
97 maxy = ty;
98 }
99 }
100 }
101 head++;
102 }
103 }
104 int main()
105 {
106 int i,startx,starty;
107 scanf("%d%d%d%d",&n,&m,&startx,&starty);
108 getchar();
109 for (i = 0;i < n;i++)
110 gets(map[i]);
111 max = fun(startx,starty);
112 maxx = startx;
113 maxy = starty;
114 book[startx][starty] = 1;
115 //dfs(startx,starty);
116 bfs(startx,starty);
117 printf("max=%d,x=%d,y=%d\n",max,maxx,maxy);
118 return 0;
119 }
四、宝岛探险
问题一
小哼降落后，所在岛屿的面积大小
输入地图和降落位置（从序号1开始）
输出降落所在位置的岛屿面积
例子
输入：
10 10 6 8
1 2 1 0 0 0 0 0 2 3
3 0 2 0 1 2 1 0 1 2
4 0 1 0 1 2 3 2 0 1
3 2 0 0 0 1 2 4 0 0
0 0 0 0 0 0 1 5 3 0
0 1 2 1 0 1 5 4 3 0
0 1 2 3 1 3 6 2 1 0
0 0 3 4 8 9 7 5 0 0
0 0 0 3 7 8 6 0 1 2
0 0 0 0 0 0 0 0 1 0
输出：
38
问题二
小哼降落后，所在岛屿的面积大小
输入地图和降落位置（从序号1开始）
输出降落所在位置的岛屿
例子
输入：
10 10 6 8
1 2 1 0 0 0 0 0 2 3
3 0 2 0 1 2 1 0 1 2
4 0 1 0 1 2 3 2 0 1
3 2 0 0 0 1 2 4 0 0
0 0 0 0 0 0 1 5 3 0
0 1 2 1 0 1 5 4 3 0
0 1 2 3 1 3 6 2 1 0
0 0 3 4 8 9 7 5 0 0
0 0 0 3 7 8 6 0 1 2
0 0 0 0 0 0 0 0 1 0
输出：
1 2 1 0 0 0 0 0 2 3
3 0 2 0 -1 -1 -1 0 1 2
4 0 1 0 -1 -1 -1 -1 0 1
3 2 0 0 0 -1 -1 -1 0 0
0 0 0 0 0 0 -1 -1 -1 0
0 -1 -1 -1 0 -1 -1 -1 -1 0
0 -1 -1 -1 -1 -1 -1 -1 -1 0
0 0 -1 -1 -1 -1 -1 -1 0 0
0 0 0 -1 -1 -1 -1 0 1 2
0 0 0 0 0 0 0 0 1 0
问题三
求出地图中小岛个数
例子
输入：
10 10
1 2 1 0 0 0 0 0 2 3
3 0 2 0 1 2 1 0 1 2
4 0 1 0 1 2 3 2 0 1
3 2 0 0 0 1 2 4 0 0
0 0 0 0 0 0 1 5 3 0
0 1 2 1 0 1 5 4 3 0
0 1 2 3 1 3 6 2 1 0
0 0 3 4 8 9 7 5 0 0
0 0 0 3 7 8 6 0 1 2
0 0 0 0 0 0 0 0 1 0
输出：
-1 -1 -1 0 0 0 0 0 -2 -2
-1 0 -1 0 -3 -3 -3 0 -2 -2
-1 0 -1 0 -3 -3 -3 -3 0 -2
-1 -1 0 0 0 -3 -3 -3 0 0
0 0 0 0 0 0 -3 -3 -3 0
0 -3 -3 -3 0 -3 -3 -3 -3 0
0 -3 -3 -3 -3 -3 -3 -3 -3 0
0 0 -3 -3 -3 -3 -3 -3 0 0
0 0 0 -3 -3 -3 -3 0 -4 -4
0 0 0 0 0 0 0 0 -4 0
五、