题目描述
注意中途返回的情况。
C++
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef pair<int, int> PII;
const int N = 110;
int n, m;
int g[N][N];
bool nxt(PII t,queue<PII>& q){
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
for (int i = 0; i < 4; i ++ ){
int x = t.first + dx[i], y = t.second + dy[i];
if (x >= 0 && x < n && y >= 0 && y < m && g[x][y] == 0 )
{
if(x==n-1&&y==m-1) return true;//到达终点提前返回
g[x][y]=1;
q.push({x, y});
}
}
return false;
}
int bfs()
{
int ans=0;
queue<PII> q;
q.push({0, 0});
g[0][0]=1;
while (q.size())
{
auto cur = q.front();q.pop();
auto end_ = q.back();
while(cur!=end_){
if(nxt(cur,q)) return ans+1;
cur = q.front();q.pop();
}
if(nxt(end_,q)) return ans+1;
ans++;//一层全访问,步数加一
}
return ans;
}
int main()
{
cin >> n >> m;
for (int i = 0; i < n; i ++ )
for (int j = 0; j < m; j ++ )
cin >> g[i][j];
cout << bfs() << endl;
return 0;
}