搜索与图论：宽度优先搜索
题目描述
输入样例
5 5
0 1 0 0 0
0 1 0 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 1 0输出样例
8参考代码
cpp
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 110;
typedef pair<int, int> PII;
int n, m;
int g[N][N]; // 存储地图
int d[N][N]; // 每一个点到起点的距离
PII q[N * N], Prev[N][N];
int bfs()
{
int hh = 0, tt = 0;
q[0] = {0, 0};
memset(d, -1, sizeof d);
d[0][0] = 0;
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
while (hh <= tt)
{
auto t = q[hh ++];
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 && d[x][y] == -1)
{
d[x][y] = d[t.first][t.second] + 1;
Prev[x][y] = t;
q[++ tt] = {x, y};
}
}
}
int x= n - 1, y = m - 1;
while (x || y)
{
cout << x << ' ' << y << endl;
auto t = Prev[x][y];
x = t.first, y = t.second;
}
return d[n - 1][m - 1];
}
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;
}