[HDU 1372] Knight Moves-蒲公英云

[HDU 1372] Knight Moves

港控/mmm° 2021-09-29 03:34 331阅读 0赞

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1372

题目描述：

在一个棋盘中，给你两个点的坐标，同时，你有八种走法，当然题目没明确说，但是这种棋盘就是这种规则。

求：从第一个点都第二个点的最少步数；

注意：

１．因为坐标是字母加数字的形式，注意下输入格式

２．如果用ｓｃａｎｆ读取输入，那么别忘了加!=EOF，否则会报 Output limit error 错误

3.　如果输入坐标的时候全部按字符格式输入的，在转成数字的时候，字母减掉 ‘a’ ，那么数字要减掉 ‘1’ ，这样横纵坐标的范围才能控制在　０－７

1 //cin字母加数字输入
 2 #include<iostream>
 3 #include<cstring>
 4 #include<queue>
 5 #include<cstdio>
 6 using namespace std;
 7 
 8 char c1,c2;
 9 int sxx,exx;
10 int sx,sy,ex,ey;
11 int go[8][2] = {
    1,2,2,1,2,-1,1,-2,-1,-2,-2,-1,-2,1,-1,2};
12 bool vis[10][10];
13 struct node
14 {
15     int x,y;
16     int count;
17 };
18 
19 bool check(node s)
20 {
21     if(s.x>=0&&s.x<8&&s.y>=0&&s.y<8)
22         return 1;
23     else
24         return 0;
25 }
26 void bfs()
27 {
28     queue<node> Q;
29     node now,nex;
30     now.x = sx;now.y = sy;
31     now.count = 0;
32     vis[now.x][now.y] = 1;
33     Q.push(now);
34     while(!Q.empty())
35     {
36         now = Q.front();
37         Q.pop();
38         if(now.x==ex&&now.y==ey)
39         {
40             printf("To get from %c%d to %c%d takes %d knight moves.\n",c1,sxx,c2,exx,now.count);
41             return ;
42         }
43         for(int i=0;i<8;i++)
44         {
45             nex.x = now.x + go[i][0];
46             nex.y = now.y + go[i][1];
47             if(check(nex)&&!vis[nex.x][nex.y])
48             {
49                 vis[nex.x][nex.y] = 1;
50                 nex.count = now.count + 1;
51                 Q.push(nex);
52             }
53         }
54     }
55     cout<<"0"<<endl;
56 }
57 
58 int main()
59 {
60     while(cin>>c1>>sxx>>c2>>exx)
61     {
62         sy = c1 - 'a';
63         sx = sxx - 1;
64         ey = c2 - 'a';
65         ex = exx - 1;
66         memset(vis,0,sizeof(vis));
67         bfs();
68     }
69     return 0;
70 }
1 //scanf字符数组输入
 2 #include<iostream>
 3 #include<cstring>
 4 #include<queue>
 5 #include<cstdio>
 6 using namespace std;
 7 
 8 char s1[3],s2[3];
 9 int sx,sy,ex,ey;
10 int go[8][2] = {
    1,2,2,1,2,-1,1,-2,-1,-2,-2,-1,-2,1,-1,2};
11 bool vis[10][10];
12 struct node
13 {
14     int x,y;
15     int count;
16 };
17 
18 bool check(node s)
19 {
20     if(s.x>=0&&s.x<8&&s.y>=0&&s.y<8)
21         return 1;
22     else
23         return 0;
24 }
25 void bfs()
26 {
27     queue<node> Q;
28     node now,nex;
29     now.x = sx;now.y = sy;
30     now.count = 0;
31     vis[now.x][now.y] = 1;
32     Q.push(now);
33     while(!Q.empty())
34     {
35         now = Q.front();
36         Q.pop();
37         if(now.x==ex&&now.y==ey)
38         {
39             printf("To get from %s to %s takes %d knight moves.\n",s1,s2,now.count);
40             return ;
41         }
42         for(int i=0;i<8;i++)
43         {
44             nex.x = now.x + go[i][0];
45             nex.y = now.y + go[i][1];
46             if(check(nex)&&!vis[nex.x][nex.y])
47             {
48                 vis[nex.x][nex.y] = 1;
49                 nex.count = now.count + 1;
50                 Q.push(nex);
51             }
52         }
53     }
54     cout<<"0"<<endl;
55 }
56 
57 int main()
58 {
59     while(scanf("%s%s",s1,s2)!=EOF)
60     {
61         sy = s1[0] - 'a';
62         sx = s1[1] - '1';
63         ey = s2[0] - 'a';
64         ex = s2[1] - '1';
65         memset(vis,0,sizeof(vis));
66         bfs();
67         getchar();
68     }
69     return 0;
70 }