没有权值考虑的广搜，不过大神用的神搜代码很简洁：
#include<cstdio>
#define min(x,y) x>y?y:x
int maze[9][9]={1,1,1,1,1,1,1,1,1,
1,0,0,1,0,0,1,0,1,
1,0,0,1,1,0,0,0,1,
1,0,1,0,1,1,0,1,1,
1,0,0,0,0,1,0,0,1,
1,1,0,1,0,1,0,0,1,
1,1,0,1,0,1,0,0,1,
1,1,0,1,0,0,0,0,1,
1,1,1,1,1,1,1,1,1};
int a,b,c,d,m;
void dfs(int x,int y,int s){
if(maze[x][y]) return;
if(x==c&&y==d){
m=min(s,m);
return;
}
s++;
maze[x][y]=1;
dfs(x+1,y,s);
dfs(x,y+1,s);
dfs(x-1,y,s);
dfs(x,y-1,s);
maze[x][y]=0;
}
int main(){
int n;
scanf("%d",&n);
while(n--){
m=9999;
scanf("%d%d%d%d",&a,&b,&c,&d);
dfs(a,b,0);
printf("%d\n",m);
}
return 0;
}
我的广搜：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
struct q
{
int row;
int col;
int step;
};
int a,b,c,d;
int head = 1,tail =2;
q que[100];
int s1[9][9]={
1,1,1,1,1,1,1,1,1,
1,0,0,1,0,0,1,0,1,
1,0,0,1,1,0,0,0,1,
1,0,1,0,1,1,0,1,1,
1,0,0,0,0,1,0,0,1,
1,1,0,1,0,1,0,0,1,
1,1,0,1,0,1,0,0,1,
1,1,0,1,0,0,0,0,1,
1,1,1,1,1,1,1,1,1};
int bfs(int a, int b, int c, int d)
{
int s[9][9];
memcpy(s,s1,sizeof(s1));
head = 1,tail =2;
que[head].row = a;
que[head].col = b;
que[head].step = 0;
while(head < tail)
{
if(que[head].row == c && que[head].col == d )
{
return que[head].step;
}
int m = que[head].row;
int n = que[head].col;
if(s[m][n+1] == 0)
{
s[m][n+1] = 1;
que[tail].row = m,que[tail].col = n+1,que[tail].step = que[head].step +1;
tail ++;
}
if(s[m+1][n] == 0)
{
s[m+1][n] = 1;
que[tail].row = m+1,que[tail].col = n,que[tail].step = que[head].step +1;
tail ++;
}
if(s[m][n-1] == 0)
{
s[m][n-1] = 1;
que[tail].row = m,que[tail].col = n-1,que[tail].step = que[head].step +1;
tail ++;
}
if(s[m-1][n] == 0)
{
s[m-1][n] =1;
que[tail].row = m-1,que[tail].col = n,que[tail].step = que[head].step +1;
tail ++;
}
head ++;
}
}
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
scanf("%d%d%d%d",&a,&b,&c,&d);
cout << bfs(a,b,c,d) << endl;
}
return 0;
}