acwing算法基础之搜索与图论--BFS

1 基础知识

BFS可以用来求取最短路，前提条件是所有边的权重一样。

2 模板

题目1：走迷宫，从左上角走到右下角，求最短路。

#include <iostream>
#include <queue>
#include <cstring>using namespace std;const int N = 110;
int d[N][N];
int g[N][N];
int n, m;int bfs() {memset(d, -1, sizeof d);queue<pair<int,int>> q;q.push({0,0});d[0][0] = 0;int dir[4][2] = {{-1,0}, {1,0}, {0,-1}, {0,1}};while (!q.empty()) {auto t = q.front();q.pop();//t下一步可以走到哪儿for (int k = 0; k < 4; ++k) {int x = t.first + dir[k][0], y = t.second + dir[k][1];if (x >= n || x < 0 || y >= m || y < 0) continue;if (g[x][y] == 1) continue;if (d[x][y] != -1) continue;d[x][y] = d[t.first][t.second] + 1;q.push({x,y});}}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;
}

扩展写法：输出最短路的路径。

#include <iostream>
#include <queue>
#include <cstring>using namespace std;const int N = 110;
int d[N][N];
int g[N][N];
int n, m;
pair<int,int> before[N][N]; //存储前一个结点int bfs() {memset(d, -1, sizeof d);queue<pair<int,int>> q;q.push({0,0});d[0][0] = 0;int dir[4][2] = {{-1,0}, {1,0}, {0,-1}, {0,1}};while (!q.empty()) {auto t = q.front();q.pop();//t下一步可以走到哪儿for (int k = 0; k < 4; ++k) {int x = t.first + dir[k][0], y = t.second + dir[k][1];if (x >= n || x < 0 || y >= m || y < 0) continue;if (g[x][y] == 1) continue;if (d[x][y] != -1) continue;d[x][y] = d[t.first][t.second] + 1;before[x][y] = t;q.push({x,y});}}//输出最短路的路径int i = n - 1, j = m - 1;while (i | j) {cout << "i = " << i << ", j = " << j << endl;pair<int,int> t = before[i][j];i = t.first;j = 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;
}