这有一个迷宫，有0~8行和0~8列：
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
0表示道路，1表示墙。
现在输入一个道路的坐标作为起点，再如输入一个道路的坐标作为终点，问最少走几步才能从起点到达终点？
（注：一步是指从一坐标点走到其上下左右相邻坐标点，如：从（3，1）到（4,1）。）
2 3 1 5 7 3 1 6 7
12 11
///Memory Limit Exceeded
#include <iostream>
#include <queue>
using namespace std;
struct node { int x, y,step; };
struct node s,e;
int dd[4][2]={-1,0,1,0,0,-1,0,1},n,ans;
int g[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
}; queue <node> q;
void bfs()
{ node t,t2;
while (! q.empty() ) q.pop();
q.push(s);
while ( !q.empty())
{ t=q.front();
q.pop();
if (t.x==e.x && t.y==e.y) { ans=t.step; return ; }
for (int i=0; i<4; i++) { t2.x=t.x+dd[i][0]; t2.y=t.y+dd[i][1]; t2.step=t.step+1;
if ( g[t2.x][t2.y]==0 ) q.push(t2);
}
}
}
int main(int argc, char *argv[])
{
cin>>n;
while (n--)
{ cin>>s.x>>s.y>>e.x>>e.y;
s.step=0;
ans=0;
bfs();
cout<<ans<<endl;
}
return 0;
}
*******************************************************************************************************
//ACCept
#include <iostream>
#include <queue>
#define N 9
#define M 9
using namespace std;
int map[N][M]= {
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 d[4][2]={-1,0,0,1,1,0,0,-1 };
struct point {
int x,y,step;
}s,e;
queue <point> my;
int BFS(point s) {
int i,x,y;
point t,temp;
while (!my.empty()) my.pop();
my.push(s);
while (!my.empty())
{
t=my.front();
my.pop();
for (i=0; i<4; i++)
{ x=t.x+d[i][0];
y=t.y+d[i][1];
if (x==e.x&& y==e.y) return t.step+1;
if (x>=0 && x<N && y>=0 && y<M && map[x][y]==0)
{ temp.x=x;
temp.y=y;
temp.step=t.step+1;
my.push(temp);
}
}
}
}
int main()
{
int n,x0,y0,ans;
cin>>n;
while (n--)
{ cin>>s.x>>s.y>>e.x>>e.y;
s.step=0;
if (s.x==e.x && s.y==e.y) ans=0;
else ans=BFS(s);
cout<<ans<<endl;
}
return 0;
}
*************************************************************
Accept
#include <cstring>
#include <iostream>
#include <queue>
using namespace std;
int ans; int g[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 dd[4][2]={-1,0,0,1,1,0,0,-1 };
int bz[10][10];
struct node{
int x,y,step;
};
struct node s,e;
queue <node> q;
int bfs( )
{
int i,x,y;
node t,t2;
q.push(s);
bz[s.x][s.y]=1;
while (!q.empty())
{ t=q.front();
q.pop();
if (t.x==e.x && t.y==e.y) return t.step;
for (i=0; i<4; i++)
{ x=t.x+dd[i][0];
y=t.y+dd[i][1];
if (g[x][y]==0&&!bz[x][y])
{ t2.x=x; t2.y=y; t2.step=t.step+1;
bz[x][y]=1;
q.push(t2);
}
}
}
}
int main()
{
int n;
cin>>n;
while (n--)
{
while (!q.empty()) q.pop();
memset(bz,0,sizeof(bz));
cin>>s.x>>s.y>>e.x>>e.y;
s.step=0;
cout<<bfs( )<<endl;
}
return 0;
}
原文地址：http://www.cnblogs.com/2014acm/p/3891176.html
