【搜索】DFS之连通性模型 + 搜索顺序
AcWing 1112. 迷宫
输入样例：
2 3 .## ..# #.. 0 0 2 2 5 ..... ###.# ..#.. ###.. ...#. 0 0 4 0输出样例:
YES NO
#include#include #include using namespace std; const int N = 110; int n; char g[N][N]; bool st[N][N]; int xa, ya, xb, yb; int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1}; bool dfs(int x, int y) { if(g[x][y] == '#') return false; if(x == xb && y == yb) return true; st[x][y] = true; for(int i = 0; i < 4; i ++ ) { int a = x + dx[i], b = y + dy[i]; if(a < 0 || a >= n || b < 0 || b >= n) continue; if(st[a][b]) continue; if(g[a][b] == '#') continue; if(dfs(a, b)) return true; } return false; } int main() { int T; cin >> T; while(T -- ) { cin >> n; for(int i = 0; i < n; i ++ ) cin >> g[i]; memset(st, 0, sizeof st); cin >> xa >> ya >> xb >> yb; if(dfs(xa, ya)) puts("YES"); else puts("NO"); } return 0; }
AcWing 1113. 红与黑
输入样例：
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 0 0输出样例：
45
#include
#include
#include
using namespace std;
const int N = 25;
int n, m;
char g[N][N];
bool st[N][N];
int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
int dfs(int x, int y)
{
int cnt = 1;
st[x][y] = true;
for(int i = 0; i < 4; i ++ )
{
int a = x + dx[i], b = y + dy[i];
if(a < 0 || a >= n || b < 0 || b >= m) continue;
if(g[a][b] != '.') continue;
if(st[a][b]) continue;
cnt += dfs(a, b);
}
return cnt;
}
int main()
{
while(cin >> m >> n, m || n)
{
for(int i = 0; i < n; i ++ ) cin >> g[i];
int x, y;
for(int i = 0; i < n; i ++ )
for(int j = 0; j < m; j ++ )
if(g[i][j] == '@')
{
x = i;
y = j;
}
memset(st, 0, sizeof st);
cout << dfs(x, y) << endl;
}
return 0;
} AcWing 1116. 马走日
输入样例：
1 5 4 0 0输出样例：
32
#include#include #include using namespace std; const int N = 10; int n, m; bool st[N][N]; int ans; int dx[8] = {-2, -1, 1, 2, 2, 1, -1, -2}; int dy[8] = {1, 2, 2, 1, -1, -2, -2, -1}; void dfs(int x, int y, int cnt) { if (cnt == n * m) { ans ++ ; return; } st[x][y] = true; for (int i = 0; i < 8; i ++ ) { int a = x + dx[i], b = y + dy[i]; if (a < 0 || a >= n || b < 0 || b >= m) continue; if (st[a][b]) continue; dfs(a, b, cnt + 1); } st[x][y] = false; } int main() { int T; scanf("%d", &T); while (T -- ) { int x, y; scanf("%d%d%d%d", &n, &m, &x, &y); memset(st, 0, sizeof st); ans = 0; dfs(x, y, 1); printf("%d\n", ans); } return 0; }
AcWing 1117. 单词接龙
输入样例：
5 at touch cheat choose tact a输出样例：
23提示
连成的“龙”为 atoucheatactactouchoose。
#include
#include
#include
using namespace std;
const int N = 21;
int n;
string word[N];
int g[N][N];
int used[N];
int ans;
void dfs(string dragon, int last)
{
ans = max((int)dragon.size(), ans);
used[last] ++ ;
for (int i = 0; i < n; i ++ )
if (g[last][i] && used[i] < 2)
dfs(dragon + word[i].substr(g[last][i]), i);
used[last] -- ;
}
int main()
{
cin >> n;
for (int i = 0; i < n; i ++ ) cin >> word[i];
char start;
cin >> start;
for (int i = 0; i < n; i ++ )
for (int j = 0; j < n; j ++ )
{
string a = word[i], b = word[j];
for (int k = 1; k < min(a.size(), b.size()); k ++ )
if (a.substr(a.size() - k, k) == b.substr(0, k))
{
g[i][j] = k;
break;
}
}
for (int i = 0; i < n; i ++ )
if (word[i][0] == start)
dfs(word[i], i);
cout << ans << endl;
return 0;
}
AcWing 1118. 分成互质组
输入样例：
6 14 20 33 117 143 175输出样例：
3
- 把某个数加到最后一组中
- 新开一个组
按照下标的顺序依次把所有的数字加入组中
#include
#include
#include
using namespace std;
const int N = 10;
int n;
int p[N];
int group[N][N];
bool st[N];
int ans = N;
int gcd(int a, int b)
{
return b ? gcd(b, a % b) : a;
}
bool check(int group[], int gc, int i)
{
for (int j = 0; j < gc; j ++ )
if (gcd(p[group[j]], p[i]) > 1)
return false;
return true;
}
void dfs(int g, int gc, int tc, int start)
{
if (g >= ans) return;
if (tc == n) ans = g;
bool flag = true;
for (int i = start; i < n; i ++ )
if (!st[i] && check(group[g], gc, i))
{
st[i] = true;
group[g][gc] = i;
dfs(g, gc + 1, tc + 1, i + 1);
st[i] = false;
flag = false;
}
if (flag) dfs(g + 1, 0, tc, 0);
}
int main()
{
cin >> n;
for (int i = 0; i < n; i ++ ) cin >> p[i];
dfs(1, 0, 0, 0);
cout << ans << endl;
return 0;
}