算法提高之搜索：多源BFS、最小步数模型、双端队列广搜

1、矩阵距离2、魔板3、电路维修

#include 
#include 
#include 

#define x first
#define y second

using namespace std;

typedef pair PII;

const int N = 1010, M = N * N;

int n, m;
char g[N][N];
PII q[M];
int dist[N][N];

void bfs()
{
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

    memset(dist, -1, sizeof dist);

    int hh = 0, tt = -1;
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
            if (g[i][j] == '1')
            {
                dist[i][j] = 0;
                q[ ++ tt] = {i, j};
            }

    while (hh <= tt)
    {
        auto t = q[hh ++ ];

        for (int i = 0; i < 4; i ++ )
        {
            int a = t.x + dx[i], b = t.y + dy[i];
            if (a < 1 || a > n || b < 1 || b > m) continue;
            if (dist[a][b] != -1) continue;

            dist[a][b] = dist[t.x][t.y] + 1;
            q[ ++ tt] = {a, b};
        }
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i ++ ) scanf("%s", g[i] + 1);

    bfs();

    for (int i = 1; i <= n; i ++ )
    {
        for (int j = 1; j <= m; j ++ ) printf("%d ", dist[i][j]);
        puts("");
    }

    return 0;
}

#include 
#include 
#include 
#include 
#include 

using namespace std;

//涉及到get和set 使用char方便操作
char g[2][4];
unordered_map> pre;
unordered_map dist;

void set(string state) 
{
    for (int i = 0; i < 4; i ++ ) g[0][i] = state[i];
    for (int i = 7, j = 0; j < 4; i --, j ++ ) g[1][j] = state[i];
}

string get()
{
    string res;
    for (int i = 0; i < 4; i ++ ) res += g[0][i];
    for (int i = 3; i >= 0; i -- ) res += g[1][i];
    return res;
}

string move0(string state)
{
    set(state);
    for (int i = 0; i < 4; i ++ ) swap(g[0][i], g[1][i]);
    return get();
}
 
string move1(string state)
{
    set(state);
    int v0 = g[0][3], v1 = g[1][3];
    for (int i = 3; i >= 0; i -- )
    {
        g[0][i] = g[0][i - 1];
        g[1][i] = g[1][i - 1];
    }
    g[0][0] = v0, g[1][0] = v1;
    return get();
}

string move2(string state)
{
    set(state);
    char v = g[0][1];
    g[0][1] = g[1][1];
    g[1][1] = g[1][2];
    g[1][2] = g[0][2];
    g[0][2] = v;
    return get();
}

int bfs(string start, string end)
{
    if (start == end) return 0;

    queue q;
    q.push(start);
    dist[start] = 0;

    while (!q.empty())
    {
        auto t = q.front();
        q.pop();

        string m[3];
        m[0] = move0(t);
        m[1] = move1(t);
        m[2] = move2(t);

        for (int i = 0; i < 3; i ++ )
            if (!dist.count(m[i]))
            {
                dist[m[i]] = dist[t] + 1;
                pre[m[i]] = {'A' + i, t};
                q.push(m[i]);
                if (m[i] == end) return dist[end];
            }
    }

    return -1;
}

int main()
{
    int x;
    string start, end;
    for (int i = 0; i < 8; i ++ )
    {
        cin >> x;
        end += char(x + '0');
    }

    for (int i = 1; i <= 8; i ++ ) start += char('0' + i);

    int step = bfs(start, end);

    cout << step << endl;

    string res;
    while (end != start)
    {
        res += pre[end].first;
        end = pre[end].second;
    }

    reverse(res.begin(), res.end());

    if (step > 0) cout << res << endl;

    return 0;
}

dijkstra可以做

点的坐标



字符的坐标


int bfs()
{
deque q;
}

#include 
#include 
#include 
#include 

#define x first
#define y second

using namespace std;

typedef pair PII;

const int N = 510, M = N * N;

int n, m;
char g[N][N];
//因为是dijsktra算法
//需要距离 和判重数组
int dist[N][N];
//判重数组是，防止使用一个点多次更新到其它点的距离
//模仿dijsktra
bool st[N][N];

int bfs()
{
    memset(dist, 0x3f, sizeof dist);
    memset(st, 0, sizeof st);
    dist[0][0] = 0;
    deque q;
    q.push_back({0, 0});
 	//按字符坐标 得到格子中的字符
 	// \转义字符 需要打2个
    char cs[] = "\/\/";
    //周围的点
    int dx[4] = {-1, -1, 1, 1}, dy[4] = {-1, 1, 1, -1};
    //周围格子中的字符坐标
    int ix[4] = {-1, -1, 0, 0}, iy[4] =  {-1, 0, 0, -1};

    while (q.size())
    {
        PII t = q.front();
        q.pop_front();
		//模仿在dijsktra中 扩展一个点到周围点的最小距离
		//出队时，这个点的dist是最小值
        if (st[t.x][t.y]) continue;
        st[t.x][t.y] = true;

        for (int i = 0; i < 4; i ++ )
        {
            int a = t.x + dx[i], b = t.y + dy[i];
            //格点的长度比格子的长度多1.
            //格子的长度是(0,0)-(n-1,m-1)
            //格点是（0,0）-(n,m)
            if (a < 0 || a > n || b < 0 || b > m) continue;
			//格子中字符
            int ca = t. x + ix[i], cb = t.y + iy[i];
   //g[ca][cb]真实格子中的字符 cs[i] 以t这个点向a，b走，需要格子字符的走向
   //相同 不加，不同则+1
            int d = dist[t.x][t.y] + (g[ca][cb] != cs[i]);

            if (d