P1101单词方阵
作者：互联网
一.题目描述：
二.解题思路：
可以先建立一个数组存储要匹配的字符串“yizhong"，然后八个方向搜索就行了。
三.代码实现：
1 #include "bits/stdc++.h"
2 using namespace std;
3 int mp[110][110];
4 char cs[110][110];
5 int road[10][2];
6 char s[10]= {'y','i','z','h','o','n','g'};
7 int n;
8 void xdfs(int sx,int sy,int f)
9 {
10 if(cs[sx][sy] != s[f]) return;
11 int dx = sx + 1;
12 int dy = sy + 0;
13 road[f][0] = sx;
14 road[f][1] = sy;
15 if(f == 6){
16 for(int i = 0;i < 7;i++)
17 mp[road[i][0]][road[i][1]] = 1;
18 return;
19 }
20 if(dx < 0 || dx >= n || dy < 0 || dy >= n) return;
21 xdfs(dx,dy,f+1);
22 }
23
24 void sdfs(int sx,int sy,int f)
25 {
26 if(cs[sx][sy] != s[f]) return;
27 road[f][0] = sx;
28 road[f][1] = sy;
29 int dx = sx - 1;
30 int dy = sy + 0;
31 if(f == 6){
32 for(int i = 0;i < 7;i++)
33 mp[road[i][0]][road[i][1]] = 1;
34 return;
35 }
36 if(dx < 0 || dx >= n || dy < 0 || dy >= n) return;
37 sdfs(dx,dy,f+1);
38 }
39 void zdfs(int sx,int sy,int f)
40 {
41 if(cs[sx][sy] != s[f]) return;
42 road[f][0] = sx;
43 road[f][1] = sy;
44 if(f == 6){
45 for(int i = 0;i < 7;i++)
46 mp[road[i][0]][road[i][1]] = 1;
47 return;
48 }
49 int dx = sx + 0;
50 int dy = sy - 1;
51 if(dx < 0 || dx >= n || dy < 0 || dy >= n) return;
52 zdfs(dx,dy,f+1);
53 }
54 void ydfs(int sx,int sy,int f)
55 {
56 if(cs[sx][sy] != s[f]) return;
57 road[f][0] = sx;
58 road[f][1] = sy;
59 if(f == 6){
60 for(int i = 0;i < 7;i++)
61 mp[road[i][0]][road[i][1]] = 1;
62 return;
63 }
64 int dx = sx + 0;
65 int dy = sy + 1;
66 if(dx < 0 || dx >= n || dy < 0 || dy >= n) return;
67 ydfs(dx,dy,f+1);
68 }
69 void xsdfs(int sx,int sy,int f)
70 {
71 if(cs[sx][sy] != s[f]) return;
72 road[f][0] = sx;
73 road[f][1] = sy;
74 int dx = sx - 1;
75 if(f == 6){
76 for(int i = 0;i < 7;i++)
77 mp[road[i][0]][road[i][1]] = 1;
78 return;
79 }
80 int dy = sy + 1;
81 if(dx < 0 || dx >= n || dy < 0 || dy >= n) return;
82 xsdfs(dx,dy,f+1);
83 }
84 void xxdfs(int sx,int sy,int f)
85 {
86 if(cs[sx][sy] != s[f]) return;
87 road[f][0] = sx;
88 road[f][1] = sy;
89 int dx = sx + 1;
90 if(f == 6){
91 for(int i = 0;i < 7;i++)
92 mp[road[i][0]][road[i][1]] = 1;
93 return;
94 }
95 int dy = sy - 1;
96 if(dx < 0 || dx >= n || dy < 0 || dy >= n) return;
97 xxdfs(dx,dy,f+1);
98 }
99 void xzdfs(int sx,int sy,int f)
100 {
101 if(cs[sx][sy] != s[f]) return;
102 road[f][0] = sx;
103 road[f][1] = sy;
104 if(f == 6){
105 for(int i = 0;i < 7;i++)
106 mp[road[i][0]][road[i][1]] = 1;
107 return;
108 }
109 int dx = sx - 1;
110 int dy = sy - 1;
111 if(dx < 0 || dx >= n || dy < 0 || dy >= n) return;
112 xzdfs(dx,dy,f+1);
113 }
114 void xydfs(int sx,int sy,int f)
115 {
116 if(cs[sx][sy] != s[f]) return;
117 road[f][0] = sx;
118 road[f][1] = sy;
119 if(f == 6){
120 for(int i = 0;i < 7;i++)
121 mp[road[i][0]][road[i][1]] = 1;
122 return;
123 }
124 int dx = sx + 1;
125 int dy = sy + 1;
126 if(dx < 0 || dx >= n || dy < 0 || dy >= n) return;
127 xydfs(dx,dy,f+1);
128 }
129 int main()
130 {
131 cin >> n;
132 getchar();
133 for(int i = 0;i < n;i++){
134 for(int j = 0;j < n;j++)
135 cin >> cs[i][j];
136 getchar();
137 }
138 for(int i = 0;i < n;i++)
139 for(int j = 0;j < n;j++){
140 xydfs(i,j,0);
141 xzdfs(i,j,0);
142 xsdfs(i,j,0);
143 xxdfs(i,j,0);
144 sdfs(i,j,0);
145 xdfs(i,j,0);
146 zdfs(i,j,0);
147 ydfs(i,j,0);
148 }
149 for(int i = 0;i < n;i++){
150 for(int j = 0;j < n;j++)
151 if(mp[i][j]) cout << cs[i][j];
152 else cout << '*';
153 cout << endl;
154 }
155 }
标签：sy,sx,P1101,int,单词,dx,dy,方阵,road 来源： https://www.cnblogs.com/scannerkk/p/15782201.html