LOJ P10028 Knight Moves 题解
每日一题 day66 打卡
我很震惊一本通居然还有这么善良的题目。
Analysis
只需要一个广搜的模板就好了，值得注意的是做每个子任务之前记得清空队列。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<queue> 6 #define int long long 7 #define maxn 310 8 #define rep(i,s,e) for(register int i=s;i<=e;++i) 9 #define dwn(i,s,e) for(register int i=s;i>=e;--i) 10 using namespace std; 11 inline int read() 12 { 13 int x=0,f=1; 14 char c=getchar(); 15 while(c<'0'||c>'9') {if(c=='-') x=-x; c=getchar();} 16 while(c>='0'&&c<='9') {x=x*10+c-'0'; c=getchar();} 17 return f*x; 18 } 19 inline void write(int x) 20 { 21 if(x<0){putchar('-');x=-x;} 22 if(x>9)write(x/10); 23 putchar(x%10+'0'); 24 } 25 int T; 26 int n; 27 int ex,ey; 28 int book[maxn][maxn]; 29 struct node 30 { 31 int x,y,step; 32 }x[maxn]; 33 queue<node> q; 34 int dx[9]={0,1,1,-1,-1,2,2,-2,-2}; 35 int dy[9]={0,2,-2,2,-2,1,-1,1,-1}; 36 signed main() 37 { 38 T=read(); 39 while(T--) 40 { 41 while(!q.empty()) q.pop(); 42 memset(book,0,sizeof(book)); 43 n=read(); 44 x[1].x=read();x[1].y=read(); 45 x[1].step=0; 46 ex=read();ey=read(); 47 book[x[1].x][x[1].y]=1; 48 if(x[1].x==ex&&x[1].y==ey) 49 { 50 write(0); 51 putchar('\n'); 52 continue; 53 } 54 q.push(x[1]); 55 while(!q.empty()) 56 { 57 node now=q.front(); 58 q.pop(); 59 int nx=now.x,ny=now.y,ns=now.step; 60 if(nx==ex&&ny==ey) 61 { 62 write(ns); 63 putchar('\n'); 64 break; 65 } 66 rep(i,1,8) 67 { 68 int xx=nx+dx[i],yy=ny+dy[i]; 69 if(book[xx][yy]==1||xx<0||xx>=n||yy<0||yy>=n) continue; 70 book[xx][yy]=1; 71 node sub; 72 sub.x=xx;sub.y=yy;sub.step=ns+1; 73 q.push(sub); 74 } 75 } 76 } 77 return 0; 78 }
如有失误请各位大佬斧正（反正我不认识斧正是什么意思）
