递推与递归 DFS深度优先搜索

Thu, 27 Feb 2025 11:30:18 +0800

原题单链接：https://www.luogu.com/paste/vjf2z3hi || https://rentry.org/5yhk52ow

例题

跳台阶

#include 
using namespace std;
const int N=1e5;
int a[N];

int f(int n)
{
 if(a[n])return a[n];
 if(n==1)return 1;
 if(n==2)return 2;
 a[n]=f(n-1)+f(n-2);
 return a[n];
}

int main()
{
 int n;
 cin>>n;
 cout
using namespace std;

const int N=20;
int st[N];
int maxn;

void dfs(int n) // n 表示当前走到第n个位置，1表示不选，2表示选，0表示还没选
{
 if(n>maxn){
 for(int i=1;i>maxn;
 dfs(1);
 return 0;
}

递归实现排列型枚举

#include 
using namespace std;

const int N=20;

int n;
bool st[N];
int arr[N];

void dfs(int x)
{
 if(x>n){
 for(int i=1;i>n;
 dfs(1);
 return 0;
}

递归实现组合型枚举

#include 
using namespace std;

const int N=25;

int n,r;
int arr[N];

void dfs(int x,int start)
{
 if(x>r){
 for(int i=1;i>n>>r;
 dfs(1,1);
 return 0;
}

P1036 选数

#include 
using namespace std;

const int N=25;

int n,k;
int arr[N],xn[N];
int ans=0;

bool isprime(int x)
{
 if(xk)
 {
 int sum=0;
 for (int i = 1; i >n>>k;
 for(int i=1;i>xn[i];
 }
 dfs(1,1);
 cout
using namespace std;

const int N = 20;

int n;
int arr[N];
int ans = 0;
vector> ansarr;

void dfs(int x, int sum)
{
 if (sum > n)
 {
 return;
 }

 if (x > 10)
 {
 if (sum == n)
 {
 ans++;
 vector temp;
 for (int i = 1; i > n;
 dfs(1, 0);
 cout 
using namespace std;

const int N = 6;

int n,m;
int arr[N],mars[N],res=0;
bool st[N];

void dfs(int x)
{
 if(x>n)
 {
 res++;
 if(res==m+1)
 for (int i = 1; i > n >> m;
 for(int i = 1; i > mars[i];
 }
 dfs(1);
 return 0;
}

P1149 火柴棒等式指数 + 预处理

#include 
using namespace std;

int n;
int num[10010] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
int a[5];
int ans = 0;

int col(int x){
 if(num[x]){
 return num[x];
 }
 else{
 int tempsum = 0;
 while(x){
 tempsum += num[x % 10];
 x /= 10;
 }
 return tempsum;
 }
}

void dfs(int x,int sum)
{
 if (sum > n-4)
 {
 return;
 }
 if (x > 3)
 {
 if (a[1] + a[2] == a[3] && sum == n - 4)
 {
 ans++;
 }
 return;
 }

 for (int i = 0; i > n;
 for (int i = 10; i 
using namespace std;

int n;
int s[15];
int b[15];
int st[15];
int ans = 1e9;

void dfs(int x)
{
 if (x > n)
 {
 bool has_tl = false;
 int sums = 1;
 int sumb = 0;
 for (int i = 1; i > n;
 for (int i = 1; i > s[i] >> b[i];
 }
 dfs(1);
 cout 
using namespace std;

int n, a, b;
int k[210];
bool st[210];
int ans[210];

void dfs(int x, int cnt)
{
 // if (cnt >= ans)
 // return;
 if (x n)
 return;

 // if (x == b)
 // {
 // ans = min(ans, cnt);
 // return;
 // }

 ans[x] = cnt;
 // 上楼
 if (x + k[x] 0 && !st[x - k[x]] && cnt + 1 > n >> a >> b;
 for (int i = 1; i > k[i];
 }
 dfs(a, 0);
 if (ans[b] == 1e9)
 ans[b] = -1;
 cout 
using namespace std;

const int N = 25;
char g[N][N];
int w, h;
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};
bool vis[N][N];
int res=0;

void dfs(int a, int b)
{
 for(int i=0;i= h || y = w)continue;
 if(g[x][y] != '.')continue;
 if(vis[x][y])continue;
 vis[x][y] = true;
 res++;
 dfs(x, y);
 }
}

int main()
{
 cin >> w >> h;
 for (int i = 0; i > g[i][j];
 for (int i = 0; i 
using namespace std;

const int N = 110;
int n, m;
char g[N][N];
bool st[N][N];
int res = 0;
int dx[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy[8] = {-1, 0, 1, 1, 1, 0, -1, -1};

void dfs(int x, int y)
{
 for(int i = 0; i = n || b = m)
 {
 continue;
 }
 if(g[a][b] == 'W' && !st[a][b])
 {
 st[a][b] = true;
 dfs(a, b);
 }
 }
}

int main()
{
 cin >> n >> m;
 for (int i = 0; i > g[i][j];
 }
 }
 for (int i = 0; i 
using namespace std;

const int N = 10;
int n, k;
char g[N][N];
bool st[N];
int res = 0;

void dfs(int x, int cnt)
{
 if (cnt == k)
 {
 res++;
 return;
 }
 if (x >= n)
 return;
 for (int i = 0; i > n >> k, n > 0 && k > 0)
 {
 for (int i = 0; i > g[i];
 res = 0;
 dfs(0, 0);
 cout 
using namespace std;

const int N = 10;
int n, k;
int ans = 0;
int arr[N];

void dfs(int x, int start, int sum)
{
 if (sum > n)
 return;
 if (x > k)
 {
 if (sum == n)
 {
 ans++;
 }
 return;
 };
 for (int i = start; sum + i * (k - x + 1) > n >> k;
 dfs(1, 1, 0);
 cout 
using namespace std;

const int N = 25;
int n;
int ans = 0;
int used[N];
string s[N];
int g[N][N];

// dragon: the current dragon string
// x: the current index of the string
void dfs(string dragon, int x)
{
 ans = max(ans, (int)dragon.size());

 used[x]++;
 for(int i=0;i> n;
 for (int i = 0; i > s[i];
 }
 char start;
 cin >> start;

 for (int i = 0; i < n; i++)
 for (int j = 0; j < n; j++)
 {
 string a=s[i],b=s[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 (s[i][0] == start)
 {
 dfs(s[i], i);
 }
 }
 cout << ans << '\n';
 return 0;
}

DFS on 蛋蛋忧桑

递推与递归 DFS深度优先搜索

例题

跳台阶

递归实现排列型枚举

递归实现组合型枚举

P1036 选数

P1149 火柴棒等式 指数 + 预处理

P1149 火柴棒等式指数 + 预处理