题目链接:https://loj.ac/problem/10028
好久没写搜索题了,找一道练练手,这道题描述可能稍微有点问题,说是n*n的地图,然后发现样例中地图大小包括了0 0点和n n点,其他就没什么了,裸的广搜,主要就是构建dir数组。
AC代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define maxn 305
using namespace std;
struct Node{
int x,y,step;
}Now,Next,S,E;
int vis[maxn][maxn];
int dir[8][2] = {1,2,1,-2,-1,2,-1,-2,2,1,2,-1,-2,1,-2,-1};
int T,n;
bool Check(int x,int y){
if(x >= 0 && y >= 0 && x <= n && y <= n && vis[x][y] == 0)return true;
return false;
}
int bfs(){
memset(vis,0,sizeof(vis));
S.step = 0;
queue<Node> q;
q.push(S);
while(!q.empty()){
Now = q.front();
q.pop();
if(Now.x == E.x && Now.y == E.y){
return Now.step;
}
for(int i=0;i<8;i++){
Next.x = Now.x + dir[i][0];
Next.y = Now.y + dir[i][1];
if(Check(Next.x,Next.y)){
vis[Next.x][Next.y] = 1;
Next.step = Now.step + 1;
q.push(Next);
}
}
}
return 0;
}
int main()
{
scanf("%d",&T);
while(T--){
scanf("%d",&n);
scanf("%d%d%d%d",&S.x,&S.y,&E.x,&E.y);
if(S.x == E.x && S.y == E.y){
printf("0\n");
continue;
}
int ans = bfs();
printf("%d\n",ans);
}
return 0;
}