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 
#include 
#include 

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 
#include 
#include 

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;
}

3 工程化

暂无。。。