Acwing-----算法提高课第二章搜索（一）

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Flood Fill

可以在线性时间复杂度内找到某个点所在的连通块

1097.池塘计数

• 链接：https://www.acwing.com/problem/content/1099/

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

const int N = 1010, M = N * N;
typedef pair<int, int> PII;
#define x first
#define y second

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

void bfs(int sx, int sy) {
    int hh = 0, tt = 0;
    q[0] = {sx, sy};
    st[sx][sy] = true;
  
    while (hh <= tt) {
        PII t = q[hh++];
        for (int i = t.x - 1; i <= t.x + 1; ++i) {
            for (int j = t.y - 1; j <= t.y + 1; ++j) {
                if (i == t.x && j == t.y) continue;
                if (i < 0 || i >= n || j < 0 || j >= m) continue;
                if (g[i][j] == ‘.‘ || st[i][j]) continue;
        
                q[++tt] = {i, j};
                st[i][j] =  true;
            }
        }
    }
}


int main() {
    cin >> n >> m;
    for (int i = 0; i < n; ++i) cin >> g[i];
  
    int cnt = 0;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            if (g[i][j] == ‘W‘ && !st[i][j]) {
                bfs(i, j);
                cnt++;
            }
        }
    }
    cout << cnt << endl;
    return 0;
}

1098.城堡问题

• 链接：https://www.acwing.com/problem/content/1100/

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;
const int N = 55, M = N * N;
typedef pair<int, int> PII;
#define x first 
#define y second
int n, m, g[N][N];
PII q[M];
bool st[N][N];

int bfs(int sx, int sy) {
    int dx[4] = {0, -1, 0, 1}, dy[4] = {-1, 0, 1, 0};
  
    int hh = 0, tt = 0;
    int area = 0;
    q[0] = {sx, sy};
    st[sx][sy] = true;
  
    while (hh <= tt) {
        PII t = q[hh++];
        area++;
        for (int i = 0 ; i< 4; ++i) {
            int a = t.x + dx[i], b = t.y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= m) continue;
            if (st[a][b]) continue;
            if (g[t.x][t.y] >> i & 1) continue;
            q[++tt] = {a, b};
            st[a][b] = true;
        }
    }
    return area;
}

int main() {
    cin >> n >> m;
    for (int i = 0 ; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            cin >> g[i][j];
        }
    }
    int cnt = 0, area = 0;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            if (!st[i][j]) {
                area = max(area, bfs(i, j));
                cnt ++;
            }
        }
    }
    cout << cnt << endl;
    cout << area << endl;
}

1106.山峰和山谷

• 链接：https://www.acwing.com/problem/content/1108/

#include <iostream>
#include <algorithm>
using namespace std;

const int  N = 1010, M = N * N;
typedef pair<int, int> PII;
#define x first
#define y second
int n, h[N][N];
PII q[M];
bool st[N][N];

void bfs(int sx, int sy, bool& has_higher, bool& has_lower) {
    int hh = 0, tt = 0;
    q[0] = {sx, sy};
    st[sx][sy] = true;
    while (hh <= tt) {
        PII t = q[hh++];
  
        for (int i = t.x - 1; i <= t.x + 1; ++i) {
            for (int j = t.y - 1; j <= t.y + 1; ++j) {
                if (i == t.x && j == t.y) continue;
                if (i < 0 || i >= n || j < 0 || j >= n) continue;
                if (h[i][j] != h[t.x][t.y]) {
                    if (h[i][j] > h[t.x][t.y]) has_higher = true;
                    else has_lower = true;
                } else if (!st[i][j]) {
                    q[++tt] = {i, j};
                    st[i][j] = true;
                }
            }
        }
    }
}

int main() {
    cin >> n;
  
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            cin >> h[i][j];
        }
    }
  
    int peak = 0, valley = 0;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            if (!st[i][j]) {
                bool has_higher = false, has_lower = false;
                bfs(i, j, has_higher, has_lower);
                if (!has_higher) peak++;
                if (!has_lower) valley++;
            }
        }
    }
    cout << peak << " " << valley << endl;
    return 0;
}

1076.迷宫问题

• 链接：https://www.acwing.com/problem/content/1078/

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef pair<int, int> PII;
#define x first
#define y second

const int N = 1010, M = N * N;
int n, g[N][N];
PII q[M], pre[N][N];

void bfs(int sx, int sy) {
    int dx[4] = {-1, 0, 1, 0,}, dy[4] = {0, 1, 0, -1};
    int hh = 0, tt = 0;
    q[0] = {sx, sy};
  
    memset(pre, -1, sizeof pre);
    pre[sx][sy] = {0, 0};
    while (hh <= tt) {
        PII t = q[hh++];
        for (int i = 0; i < 4; ++i) {
            int a = t.x + dx[i], b = t.y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= n) continue;
            if (g[a][b]) continue;
            if (pre[a][b].x != -1) continue;
      
            q[++tt] = {a, b};
            pre[a][b] = t;
        }
    }
}

int main() {
    cin >> n;
    for (int i = 0; i < n; ++i) 
        for (int j = 0; j < n; ++j) 
            cin >> g[i][j];
      
    bfs(n - 1,n - 1);
    PII end(0, 0);
  
    while (true) {
        cout << end.x << " " << end.y << endl;
        if (end.x == n - 1 && end.y == n - 1) break;
        end = pre[end.x][end.y];
    }
    return 0;
}

188.武士风度的牛

• 链接：https://www.acwing.com/problem/content/190/

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
#define x first
#define y second
typedef pair<int, int> PII;

const int N = 155, M = N * N;
int n, m, dist[N][N];
char g[N][N];
PII q[M];

int bfs() {
    int dx[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
    int dy[8] = {1, 2, 2, 1, -1, -2, -2, -1};
    int sx, sy;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            if (g[i][j] == ‘K‘) {
                sx = i, sy = j;
            }
        }
    }
    int hh = 0, tt = 0;
    q[0] = {sx, sy};
    memset(dist, -1, sizeof dist);
    dist[sx][sy] = 0;
  
    while (hh <= tt) {
        PII t = q[hh++];
        for (int i = 0; i < 8; ++i) {
            int a = t.x + dx[i], b = t.y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= m) continue;
            if (g[a][b] == ‘*‘) continue;
            if (dist[a][b] != -1) continue;
      
            if (g[a][b] == ‘H‘) return dist[t.x][t.y] + 1;
      
            dist[a][b] = dist[t.x][t.y] + 1;
            q[++tt] = {a, b};
        }
    }
    return -1;
}

int main() {
    cin >> m >> n;
    for (int i = 0; i < n; ++i) {
            cin >> g[i];   
    }
  
    cout << bfs() << endl;
    return 0;
}

1100.抓住那头牛

• 链接：https://www.acwing.com/problem/content/1102/

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;
const int N = 200010;

int n, k;
int q[N];
int dist[N];

int bfs() {
    memset(dist, -1, sizeof dist);
    dist[n] = 0;
    q[0] = n;
  
    int hh = 0, tt = 0;
    while (hh <= tt) {
        int t = q[hh++];
        if (t == k) return dist[k];
        if (t + 1 < N && dist[t + 1] == -1) {
            dist[t + 1] = dist[t] + 1;
            q[++tt] = t + 1;
        }
        if (t - 1 >= 0 && dist[t - 1] == -1) {
            dist[t - 1] = dist[t] + 1;
            q[++tt] = t - 1;
        }
        if (t * 2 < N && dist[t * 2] == -1) {
            dist[t * 2] = dist[t] + 1;
            q[++tt] = t * 2;
        }
    }
    return -1;
}

int main() {
    cin >> n >> k;
    cout << bfs() << endl;
    return 0;
}

Acwing-----算法提高课第二章搜索（一）

标签：bfs   end   std   ||   class   continue   win   搜索   har   

原文地址：https://www.cnblogs.com/clown9804/p/13714416.html