Acwing-----算法提高课第二章搜索（一）
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
#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
#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
#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
#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
#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
#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;
}
