A. Salem and Sticks
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 1010 5 int n, a[N]; 6 7 int work(int x) 8 { 9 int res = 0; 10 for (int i = 1; i <= n; ++i) 11 res += max(0, abs(x - a[i]) - 1); 12 return res; 13 } 14 15 int main() 16 { 17 while (scanf("%d", &n) != EOF) 18 { 19 for (int i = 1; i <= n; ++i) scanf("%d", a + i); 20 int Min = (int)1e9, pos = -1; 21 for (int i = 1; i <= 100; ++i) 22 { 23 int tmp = work(i); 24 if (tmp < Min) 25 { 26 Min = tmp; 27 pos = i; 28 } 29 } 30 printf("%d %d ", pos, Min); 31 } 32 return 0; 33 }
B. Zuhair and Strings
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 200010 5 int n, k; 6 char s[N]; 7 8 int work(char c) 9 { 10 int res = 0; 11 int tmp = 0; 12 for (int i = 1; i <= n; ++i) 13 { 14 if (s[i] != c) tmp = 0; 15 else 16 { 17 ++tmp; 18 if (tmp == k) 19 { 20 ++res; 21 tmp = 0; 22 } 23 } 24 } 25 return res; 26 } 27 28 int main() 29 { 30 while (scanf("%d%d", &n, &k) != EOF) 31 { 32 scanf("%s", s + 1); 33 int res = 0; 34 for (int i = 'a'; i <= 'z'; ++i) 35 res = max(res, work(i)); 36 printf("%d ", res); 37 } 38 return 0; 39 }
C. Ayoub and Lost Array
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 200010 6 const ll MOD = (ll)1e9 + 7; 7 int n, l, r; 8 ll a[3], f[N][3]; 9 10 int main() 11 { 12 while (scanf("%d%d%d", &n, &l, &r) != EOF) 13 { 14 memset(a, 0, sizeof a); 15 while (l % 3 && l <= r) 16 { 17 a[l % 3]++; 18 ++l; 19 } 20 while (r % 3 && l <= r) 21 { 22 a[r % 3]++; 23 --r; 24 } 25 if (l <= r) 26 { 27 ++a[0]; 28 int tmp = (r - l) / 3; 29 a[0] += tmp; 30 a[1] += tmp; 31 a[2] += tmp; 32 } 33 memset(f, 0, sizeof f); 34 f[0][0] = 1; 35 for (int i = 1; i <= n; ++i) 36 { 37 f[i][0] = (f[i - 1][0] * a[0] % MOD + f[i - 1][1] * a[2] % MOD + f[i - 1][2] * a[1] % MOD) % MOD; 38 f[i][1] = (f[i - 1][0] * a[1] % MOD + f[i - 1][1] * a[0] % MOD + f[i - 1][2] * a[2] % MOD) % MOD; 39 f[i][2] = (f[i - 1][0] * a[2] % MOD + f[i - 1][1] * a[1] % MOD + f[i - 1][2] * a[0] % MOD) % MOD; 40 } 41 printf("%lld ", f[n][0]); 42 } 43 return 0; 44 }
D. Kilani and the Game
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 1010 5 int n, m, p; 6 char G[N][N]; 7 int s[10]; 8 struct node 9 { 10 int x, y, step; 11 node () {} 12 node (int x, int y, int step) : x(x), y(y), step(step) {} 13 }; 14 queue <node> q[10]; 15 bool stop() 16 { 17 for (int i = 1; i <= p; ++i) if (!q[i].empty()) 18 return false; 19 return true; 20 } 21 22 int Move[][2] = 23 { 24 -1, 0, 25 1, 0, 26 0,-1, 27 0, 1, 28 }; 29 bool ok(int x, int y) 30 { 31 if (x < 1 || x > n || y < 1 || y > m || G[x][y] != '.') return false; 32 return true; 33 } 34 35 void BFS(int id, int cnt) 36 { 37 while (!q[id].empty()) 38 { 39 int x = q[id].front().x; 40 int y = q[id].front().y; 41 int step = q[id].front().step; 42 //printf("%d %d %d %d ", x, y, id, step); 43 if (step / s[id] >= cnt) return; 44 q[id].pop(); 45 for (int i = 0; i < 4; ++i) 46 { 47 int nx = x + Move[i][0]; 48 int ny = y + Move[i][1]; 49 if (ok(nx, ny)) 50 { 51 G[nx][ny] = id + '0'; 52 q[id].push(node(nx, ny, step + 1)); 53 } 54 } 55 } 56 } 57 58 int main() 59 { 60 while (scanf("%d%d%d", &n, &m, &p) != EOF) 61 { 62 for (int i = 1; i <= p; ++i) scanf("%d", s + i); 63 for (int i = 1; i <= n; ++i) scanf("%s", G[i] + 1); 64 for (int i = 1; i <= p; ++i) while (!q[i].empty()) q[i].pop(); 65 for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) if (isdigit(G[i][j])) 66 q[G[i][j] - '0'].push(node(i, j, 0)); 67 int cnt = 1; 68 while (1) 69 { 70 for (int i = 1; i <= p; ++i) BFS(i, cnt); 71 ++cnt; 72 if (stop()) break; 73 } 74 int ans[10]; 75 memset(ans, 0, sizeof ans); 76 for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) if (isdigit(G[i][j])) 77 ++ans[G[i][j] - '0']; 78 //for (int i = 1; i <= n; ++i) printf("%s ", G[i] + 1); 79 for (int i = 1; i <= p; ++i) printf("%d%c", ans[i], " "[i == p]); 80 } 81 return 0; 82 }
E. Helping Hiasat
Upsolved.
题意：
两种操作
- 更改自己的handle
- 伙伴查询handle
如果一个伙伴在每次查询时显示的都是自己名字，那么他就会开心
问 最多可以让多少人开心
思路：
法一：
一张图的最大独立集是选出一个点集，使得任意两点不相邻
一张图的最大团是选出一个点集，使得任意两点之间有边相连
一张无向图的补图的最大图就是原图的最大独立集
我们发现这道题两个1之间的所有点都是不能一起happy的，那我们给他们两两之间连上边
然后求补图的最大团即可
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 110 5 int n, m, t; 6 int g[N][N]; 7 int dp[N]; 8 int stk[N][N]; 9 int mx; 10 11 map <string, int> mp; 12 int get(string s) 13 { 14 if (mp.find(s) != mp.end()) return mp[s]; 15 else mp[s] = t++; 16 return mp[s]; 17 } 18 19 int DFS(int n, int ns, int dep) 20 { 21 if (ns == 0) 22 { 23 mx = max(mx, dep); 24 return 1; 25 } 26 int i, j, k, p, cnt; 27 for (i = 0; i < ns; ++i) 28 { 29 k = stk[dep][i]; 30 cnt = 0; 31 if (dep + n - k <= mx) 32 return 0; 33 if (dep + dp[k] <= mx) 34 return 0; 35 for (j = i + 1; j < ns; ++j) 36 { 37 p = stk[dep][j]; 38 if (g[k][p]) 39 stk[dep + 1][cnt++] = p; 40 } 41 DFS(n, cnt, dep + 1); 42 } 43 return 1; 44 } 45 46 int clique(int n) 47 { 48 int i, j, ns; 49 for (mx = 0, i = n - 1; i >= 0; --i) 50 { 51 for (ns = 0, j = i + 1; j < n; ++j) 52 { 53 if (g[i][j]) 54 stk[1][ns++] = j; 55 } 56 DFS(n, ns, 1); 57 dp[i] = mx; 58 } 59 return mx; 60 } 61 62 vector <int> vec; 63 void add() 64 { 65 vec.erase(unique(vec.begin(), vec.end()), vec.end()); 66 for (auto u : vec) for (auto v : vec) 67 g[u][v] = g[v][u] = 0; 68 vec.clear(); 69 } 70 71 int main() 72 { 73 while (scanf("%d%d", &n, &m) != EOF) 74 { 75 t = 0; 76 mp.clear(); 77 for (int i = 0; i < m; ++i) for (int j = 0; j < m; ++j) g[i][j] = 1; 78 []() 79 { 80 int op; char s[50]; 81 for (int nn = 1; nn <= n; ++nn) 82 { 83 scanf("%d", &op); 84 if (op == 1) 85 add(); 86 else 87 { 88 scanf("%s", s + 1); 89 vec.push_back(get(s + 1)); 90 } 91 } 92 }(); 93 add(); 94 for (int i = 0; i < m; ++i) g[i][i] = 1; 95 //for (int i = 1; i <= m; ++i) for (int j = 1; j <= m; ++j) printf("%d %d %d ", i, j, g[i][j]); 96 printf("%d ", clique(m)); 97 } 98 return 0; 99 }
法二：
$m = 40, 可以折半状压，再合起来$
考虑妆压的时候要从子集转移到超集，可以通过$dp上去$
$vp的时候想到折半状压，但是当时是枚举子集来转移，复杂度大大增加..$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 50 5 #define M 1100010 6 int n, m, t; 7 int G[N][N]; 8 9 map <string, int> mp; 10 int get(string s) 11 { 12 if (mp.find(s) != mp.end()) return mp[s]; 13 else mp[s] = ++t; 14 return mp[s]; 15 } 16 17 vector <int> vec; 18 void add() 19 { 20 vec.erase(unique(vec.begin(), vec.end()), vec.end()); 21 for (auto u : vec) for (auto v : vec) 22 G[u][v] = 1; 23 vec.clear(); 24 } 25 26 int f[M], g[M]; 27 int main() 28 { 29 while (scanf("%d%d", &n, &m) != EOF) 30 { 31 t = 0; mp.clear(); 32 memset(G, 0, sizeof G); 33 int op; char s[N]; 34 for (int nn = 1; nn <= n; ++nn) 35 { 36 scanf("%d", &op); 37 if (op == 1) add(); 38 else 39 { 40 scanf("%s", s + 1); 41 vec.push_back(get(s + 1)); 42 } 43 } 44 add(); 45 for (int i = 1; i <= m; ++i) 46 G[i][i] = 0; 47 int s1 = m / 2, s2 = m - s1; 48 for (int i = 1; i < (1 << s1); i <<= 1) f[i] = 1; 49 for (int i = 1; i < (1 << s2); i <<= 1) g[i] = 1; 50 f[0] = g[0] = 0; 51 for (int i = 1; i < (1 << s1); ++i) 52 { 53 for (int j = 0; j < s1; ++j) if (!((i >> j) & 1)) 54 { 55 int flag = 1; 56 for (int k = 0; k < s1; ++k) if (((i >> k) & 1) && G[k + 1][j + 1]) 57 { 58 flag = 0; 59 break; 60 } 61 f[i | (1 << j)] = max(f[i | (1 << j)], f[i] + flag); 62 } 63 } 64 for (int i = 1; i < (1 << s2); ++i) 65 { 66 for (int j = 0; j < s2; ++j) if (!((i >> j) & 1)) 67 { 68 int flag = 1; 69 for (int k = 0; k < s2; ++k) if (((i >> k) & 1) && G[s1 + k + 1][s1 + j + 1]) 70 { 71 flag = 0; 72 break; 73 } 74 g[i | (1 << j)] = max(g[i | (1 << j)], g[i] + flag); 75 } 76 } 77 int res = 0; 78 for (int i = 0; i < (1 << s1); ++i) 79 { 80 int s3 = (1 << s2) - 1; 81 for (int j = 0; j < s1; ++j) if ((i >> j) & 1) 82 { 83 for (int k = 0; k < s2; ++k) if (G[j + 1][s1 + k + 1] && ((s3 >> k) & 1)) 84 s3 ^= (1 << k); 85 } 86 res = max(res, f[i] + g[s3]); 87 } 88 printf("%d ", res); 89 } 90 return 0; 91 }
法三：
为什么随机也行啊，能不能证明啊，喵喵喵..
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 50 6 int n, m, t; 7 int G[N][N], a[N]; 8 ll f[N]; 9 10 map <string, int> mp; 11 int get(string s) 12 { 13 if (mp.find(s) != mp.end()) return mp[s]; 14 else mp[s] = ++t; 15 return mp[s]; 16 } 17 18 vector <int> vec; 19 void add() 20 { 21 vec.erase(unique(vec.begin(), vec.end()), vec.end()); 22 for (auto u : vec) for (auto v : vec) 23 G[u][v] = 1; 24 vec.clear(); 25 } 26 27 int main() 28 { 29 while (scanf("%d%d", &n, &m) != EOF) 30 { 31 t = 0; mp.clear(); 32 memset(G, 0, sizeof G); 33 int op; char s[N]; 34 for (int nn = 1; nn <= n; ++nn) 35 { 36 scanf("%d", &op); 37 if (op == 1) add(); 38 else 39 { 40 scanf("%s", s + 1); 41 vec.push_back(get(s + 1)); 42 } 43 } 44 add(); 45 for (int i = 1; i <= m; ++i) 46 G[i][i] = 0; 47 int res = 0; 48 for (int i = 1; i <= m; ++i) a[i] = i; 49 for (int i = 1; i <= m; ++i) 50 { 51 f[i] = 0; 52 for (int j = 1; j <= m; ++j) if (G[i][j]) 53 f[i] |= (1ll << (j - 1)); 54 } 55 for (int t = 1; t <= 500; ++t) 56 { 57 random_shuffle(a + 1, a + 1 + m); 58 ll g = 0; int tmp = 0; 59 for (int i = 1; i <= m; ++i) if (((g >> (a[i] - 1)) & 1ll) == 0) 60 { 61 g |= f[a[i]]; 62 ++tmp; 63 } 64 res = max(res, tmp); 65 } 66 printf("%d ", res); 67 } 68 return 0; 69 }