洛谷P2089烤鸡

#include<iostream>
using namespace std;
const int N = 20, M = 1000010;
int ans[N];
int dp[M][N];
int n, count;
void dfs(int x, int sum){if(sum > n)return;if(x > 10){if(sum == n){count++;for(int i = 1; i <= n; i++)dp[count][i] = ans[i];}return;}for(int i = 1; i <= 3; i++){ans[x] = i;dfs(x + 1, sum + i);ans[x] = 0; }
} 
int main(){cin >> n;dfs(1, 0);cout << count << endl;for(int i = 1; i <= count; i++){for(int j = 1; j <= 10; j++){cout << dp[i][j] << " ";}cout << endl;}return 0;
} 

洛谷P1088火星人

next_permutation函数

#include<iostream>
#include<algorithm>
using namespace std;
int n, m;
const int N = 10010;
int a[N];int main(){cin >> n >> m;for(int i = 1; i <= n; i++)cin >> a[i];if(m == 0){for(int i = 1; i <= n; i++)cout << a[i] << " ";}while(m--){next_permutation(a + 1, a + n + 1);}for(int i = 1; i <= n; i++)cout << a[i] << " "; return 0;
}

 DFS

#include<iostream>
#include<algorithm>
using namespace std;
int n, m;
const int N = 10010;
int ans[N], a[N];
bool st[N];
int cnt;
void dfs(int x){if(cnt >= m + 1)return;if(x > n){cnt++;if(cnt == m + 1){for(int i = 1; i <= n; i++)cout << ans[i] << " ";}return;}for(int i = 1; i <= n; i++){if(!cnt){i = a[x];}if(!st[i]){ans[x] = i;st[i] = true;dfs(x + 1);ans[x] = 0;st[i] = false;}}
}
int main(){cin >> n >> m;for(int i = 1; i <= n; i++)cin >> a[i];dfs(1);return 0;
}

洛谷P1149火柴棒

#include<iostream>
#include<algorithm>
using namespace std;
int n;
const int N = 10010;
int dist[N] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
int arr[4];
int ans;
void dfs(int x, int sum){if(sum > n)return;if(x > 3){if(arr[1] + arr[2] == arr[3] && sum == n){ans++;}return;}for(int i = 0; i <= 1000; i++){arr[x] = i;dfs(x + 1, sum + dist[i]);arr[x] = 0;}
}
int main(){cin >> n;n -= 4;for(int i = 10; i <= 10000; i++){dist[i] = dist[i % 10] + dist[i / 10];}dfs(1, 0);cout << ans << endl;return 0;
}

洛谷P2036PERKET

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int n;
const int N = 20;
int acid[N], bitter[N];
int ans = 0x7fffffff;
bool st[N];void dfs(int x){if(x > n){int sum1 = 1;int sum2 = 0;bool has_sc = false;for(int i = 1; i <= n; i++){if(st[i] != 0){has_sc = true;sum1 *= acid[i];sum2 += bitter[i];}if(has_sc)ans = min(ans, abs(sum1 - sum2));}return;}st[x] = true;dfs(x + 1);st[x] = false; dfs(x + 1); 
}
int main(){cin >> n;for(int i = 1; i <= n; i++)cin >> acid[i] >> bitter[i];dfs(1);cout << ans << endl;return 0;
}

洛谷P1135奇怪的电梯

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int n, a, b;
const int N = 210;
int s[N];
int ans = 0x7fffffff;
bool st[N];
void dfs(int x, int cnt){if(cnt >= ans)return;if(x == b){ans = min(ans, cnt);return;}int up = x + s[x];if(up <= n && !st[up]){st[up] = true;dfs(x + s[x], cnt + 1);st[up] = false;}int down = x - s[x];if(down >= 1 && !st[down]){st[down] = true;dfs(x - s[x], cnt + 1);st[down] = false; }
}
int main(){ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);cin >> n >> a >> b;for(int i = 1; i <= n; i++)cin >> s[i];st[a] = true;dfs(a, 0);if(ans == 0x7fffffff)cout << -1 << endl;else cout << ans << endl;return 0;
}

洛谷P1683

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int n, m;
const int N = 25;
char mp[N][N];
bool st[N][N];
int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};
int ans = 1;
void dfs(int x, int y){for(int i = 0; i < 4; i++){int a = x + dx[i], b = y + dy[i];if(a < 1 || a > n || b < 1 || b > m)continue;if(mp[a][b] != '.')continue;if(st[a][b])continue;st[a][b] = true;ans++;dfs(a, b);}
}
int main(){ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);cin >> m >> n;for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){cin >> mp[i][j];}}for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){if(mp[i][j] == '@'){st[i][j] = true;dfs(i, j);}}}cout << ans << endl;return 0;
}

洛谷P1596

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int N = 110;
char g[N][N];
bool st[N][N];
int n, m;
int ans;
int dx[8] = {0, 0, 1, -1, 1, 1, -1, -1}, dy[8] = {1, -1, 0, 0, 1, -1, 1, -1};
void dfs(int x, int y){for(int i = 0; i < 8; i++){int a = x + dx[i], b = y + dy[i];if(a < 1 || a > n || b < 1 || b > m)continue;if(st[a][b] || g[a][b] != 'W')continue;st[a][b] = true;dfs(a, b);}
}
int main(){cin >> n >> m;for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){cin >> g[i][j];}}for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){if(g[i][j] == 'W' && !st[i][j]){ans++;st[i][j] = true;dfs(i, j);}}}cout << ans << endl; return 0;
}

acwing1114棋盘问题

#include<iostream>
#include<cstring>
using namespace std;
int n, m;
const int N = 10;
char g[N][N];
bool col[N];
int ans;
void dfs(int x, int cnt){if(cnt == m){ans++;return;}if(x > n)return;for(int i = 1; i <= n; i++){if(!col[i] && g[x][i] == '#'){col[i] = true;dfs(x + 1, cnt + 1);col[i] = false;}}dfs(x + 1, cnt);
}
int main(){while(cin >> n >> m, (n != -1 && m != -1)){memset(g, 0, sizeof(g));ans = 0;for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){cin >> g[i][j];}}memset(col, 0, sizeof(col));dfs(1, 0);cout << ans << endl;}return 0;
}

洛谷P1025数的划分

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int n, k;
const int N = 210;
int ans;
void dfs(int x, int cnt, int start){if(x > k){if(cnt == 0)ans++;return;}if(x > k || cnt < 0)return;for(int i = start; i <= cnt;  i++){dfs(x + 1, cnt - i, i);}
}
int main(){ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);cin >> n >> k;dfs(1, n, 1);cout << ans << endl;return 0;
}

洛谷P1019单词接龙

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int n;
const int N = 25;
string str[N];
int g[N][N];//g[i][j]表示第i个字符串到j个的共同部分的长度 
int used[100];
int ans;
void dfs(string dragon, int x){ans = max(ans, (int)dragon.size());used[x]++;for(int i = 1; i <= n; i++){if(g[x][i] && used[i] < 2){dfs(dragon + str[i].substr(g[x][i]), i);}}used[x]--;
}
int main(){cin >> n;for(int i = 1; i <= n; i++)cin >> str[i];char start; cin >> start;for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){for(int k = 1; k <= min(str[i].size(), str[j].size()); k++){string a = str[i], b = str[j];if(a.substr(a.size() - k, k) == b.substr(0, k)){g[i][j] = k;break;}}}}for(int i = 1; i <= n; i++){if(str[i][0] == start){dfs(str[i], i);}}cout << ans << endl;return 0;
}