我也是BFS,MLE了，0分
```
#include<bits/stdc++.h>
using namespace std;
char ma[1005][1005];
int dx[] = {0,1,0,-1,0},dy[] = {0,0,1,0,-1},n,sx,sy,fx,fy,ans = 0x3f3f3f3f;
struct p
{
int x;
int y;
int f;
int b;
};
int main()
{
cin >> n;
for(int i = 1;i <= n;i++) for(int j = 1;j <= n;j++) cin >> ma[i][j];
cin >> sx >> sy >> fx >> fy;
queue<p> q;
q.push({sx,sy,0,0});
while(!q.empty())
{
p t = q.front();
q.pop();
if(t.x < 1 || t.x > n || t.y < 1 || t.y > n || ma[t.x][t.y] == '1') continue;
if(t.x == fx && t.y == fy) ans = min(ans,t.b);
for(int i = 1;i <= 4;i++)
{
if(i == t.f) continue;
q.push({t.x+dx[i],t.y+dy[i],(i+1)%4+1,t.b+1});
}
}
cout << ans;
return 0;
}
```
by xingyu671 @ 2023-12-06 11:46:59
两个都mle了？
by lightme @ 2023-12-12 21:57:19
```
#include <bits/stdc++.h>
#define PII pair<int,int>
using namespace std;
int c[1005][1005],cnt[1005][1005];
int dy[5]={0,-1,1,0};
int dx[5]={1,0,0,-1};
int x,y,xx,yy,n;
void bfs(){
//memset(cnt,-1,sizeof cnt);
queue<PII>q;
q.push({x,y});
int x1,y1,xd,yd;
while(!q.empty()){
x1=q.front().first;
y1=q.front().second;
for(int i=0;i<4;++i){
xd=x1+dx[i];
yd=y1+dy[i];
if(xd>0&&xd<=n&&yd>0&&yd<=n&&c[xd][yd]==0&&cnt[xd][yd]==0){
//if(cnt[xd][yd]>cnt[x1][y1]||cnt[xd][yd]==0){
q.push({xd,yd});
cnt[xd][yd]=cnt[x1][y1]+1;
//}
//if(cnt[xx][yy]>0) return;
}
}
q.pop();
}
}
int main(){
cin>>n;
for(int i=1;i<=n;++i){
string a;
cin>>a;
for(int j=0;j<a.size();++j){
c[i][j+1]=a[j]-'0';
}
}
cin>>x>>y>>xx>>yy;
bfs();
cout<<cnt[xx][yy];
}
```
按照第一个bfs改的
by lightme @ 2023-12-12 22:09:56
```
if(cnt[xd][yd]>cnt[x1][y1]||cnt[xd][yd]==0)
```
这一句有点多余了因为后面的cnt下一个cnt都是大于前一个的按照bfs的特点来说，所以不需要做一步，估计你过的那个数据点就正好是样例，数据卡好的正好是最后一个数据所以无所谓。后面数据量大了就有问题了这里爆内存的估计是队列爆的，不是这个二维数组所以不用像下一个bfs一样做个结构体。小tip可以后面加个return这样稍微节约一点时间
by lightme @ 2023-12-12 22:13:19
同样bfs，mle了:(
```cpp
#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cstdlib>
using namespace std;
int n;
int map[1005][1005];
char a[1005][1005];
int x11,y11,x22,y22;
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
bool vis[1005][1005];
struct node
{
int x,y,count;
}fir;
queue <node>q;
int main()
{
cin>>n;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
cin>>a[i][j];
map[i][j]=a[i][j]-'0';
}
}
cin>>x11>>y11>>x22>>y22;
fir.x=x11;
fir.y=y11;
vis[x11][y11]=1;
fir.count=0;
q.push(fir);
struct node top;
while(q.empty()==0)
{
top=q.front();//cout<<top.x<<" "<<top.y<<endl;
q.pop();
if(top.x==x22&&top.y==y22)goto what;
for(int i=0;i<4;i++)
{
struct node nxt;
nxt.x=top.x+dx[i];
nxt.y=top.y+dy[i];
if(nxt.x<1||nxt.x>n||nxt.y<1||nxt.y>n)continue;
if(map[nxt.x][nxt.y]==1)continue;
if(vis[nxt.x][nxt.y]==1)continue;
nxt.count=top.count+1;
vis[top.x][top.y]=1;
q.push(nxt);
}
}
what:
cout<<top.count;
return 0;
}
by ARTI001 @ 2024-01-04 11:59:13