Codeforces Round #533 (Div. 2) Solution

A. Salem and Sticks

签.

 #include <bits/stdc++.h>

 using namespace std;

 #define N 1010

 int n, a[N];

 int work(int x)

 {

     int res = ;

     for (int i = ; i <= n; ++i)

         res += max(, abs(x - a[i]) - );

     return res;

 }

 int main()

 {

     while (scanf("%d", &n) != EOF)

     {

         for (int i = ; i <= n; ++i) scanf("%d", a + i);

         int Min = (int)1e9, pos = -;

         for (int i = ; i <= ; ++i)

         {

             int tmp = work(i);

             if (tmp < Min)

             {

                 Min = tmp;

                 pos = i;

             }

         }

         printf("%d %d\n", pos, Min);

     }

     return ;

 }

B. Zuhair and Strings

签.

 #include <bits/stdc++.h>

 using namespace std;

 #define N 200010

 int n, k;

 char s[N];

 int work(char c)

 {

     int res = ;

     int tmp = ;

     for (int i = ; i <= n; ++i)

     {

         if (s[i] != c) tmp = ;

         else

         {

             ++tmp;

             if (tmp == k)

             {

                 ++res;

                 tmp = ;

             }

         }

     }

     return res;

 }

 int main()

 {

     while (scanf("%d%d", &n, &k) != EOF)

     {

         scanf("%s", s + );

         int res = ;

         for (int i = 'a'; i <= 'z'; ++i)

             res = max(res, work(i));

         printf("%d\n", res);

     }

     return ;

 }

C. Ayoub and Lost Array

签.

 #include <bits/stdc++.h>

 using namespace std;

 #define ll long long

 #define N 200010

 const ll MOD = (ll)1e9 + ;

 int n, l, r;

 ll a[], f[N][]; 

 int main()

 {

     while (scanf("%d%d%d", &n, &l, &r) != EOF)

     {

         memset(a, , sizeof a);

         while (l %  && l <= r)

         {

             a[l % ]++;

             ++l;

         }

         while (r %  && l <= r)

         {

             a[r % ]++;

             --r;

         }

         if (l <= r)

         {

             ++a[];

             int tmp = (r - l) / ;

             a[] += tmp;

             a[] += tmp;

             a[] += tmp;

         }

         memset(f, , sizeof f);

         f[][] = ;

         for (int i = ; i <= n; ++i)

         {

             f[i][] = (f[i - ][] * a[] % MOD + f[i - ][] * a[] % MOD + f[i - ][] * a[] % MOD) % MOD;

             f[i][] = (f[i - ][] * a[] % MOD + f[i - ][] * a[] % MOD + f[i - ][] * a[] % MOD) % MOD;

             f[i][] = (f[i - ][] * a[] % MOD + f[i - ][] * a[] % MOD + f[i - ][] * a[] % MOD) % MOD;

         }

         printf("%lld\n", f[n][]);

     }

     return ;

 }

D. Kilani and the Game

签.

 #include <bits/stdc++.h>

 using namespace std;

 #define N 1010

 int n, m, p;

 char G[N][N];

 int s[];

 struct node

 {

     int x, y, step;

     node () {}

     node (int x, int y, int step) : x(x), y(y), step(step) {}

 };

 queue <node> q[];

 bool stop()

 {

     for (int i = ; i <= p; ++i) if (!q[i].empty())

         return false;

     return true;

 }

 int Move[][] =

 {

     -, ,

      , ,

      ,-,

      , ,

 };

 bool ok(int x, int y)

 {

     if (x <  || x > n || y <  || y > m || G[x][y] != '.') return false;

     return true;

 }

 void BFS(int id, int cnt)

 {

     while (!q[id].empty())

     {

         int x = q[id].front().x;

         int y = q[id].front().y;

         int step = q[id].front().step;

         //printf("%d %d %d %d\n", x, y, id, step);

         if (step / s[id] >= cnt) return;

         q[id].pop();

         for (int i = ; i < ; ++i)

         {

             int nx = x + Move[i][];

             int ny = y + Move[i][];

             if (ok(nx, ny))

             {

                 G[nx][ny] = id + '';

                 q[id].push(node(nx, ny, step + ));

             }

         }

     }

 }    

 int main()

 {

     while (scanf("%d%d%d", &n, &m, &p) != EOF)

     {

         for (int i = ; i <= p; ++i) scanf("%d", s + i);

         for (int i = ; i <= n; ++i) scanf("%s", G[i] + );

         for (int i = ; i <= p; ++i) while (!q[i].empty()) q[i].pop();

         for (int i = ; i <= n; ++i) for (int j = ; j <= m; ++j) if (isdigit(G[i][j]))

             q[G[i][j] - ''].push(node(i, j, ));

         int cnt = ;

         while ()

         {

             for (int i = ; i <= p; ++i) BFS(i, cnt);

             ++cnt;

             if (stop()) break;

         }

         int ans[];

         memset(ans, , sizeof ans);

         for (int i = ; i <= n; ++i) for (int j = ; j <= m; ++j) if (isdigit(G[i][j]))

             ++ans[G[i][j] - ''];

         //for (int i = 1; i <= n; ++i) printf("%s\n", G[i] + 1);

         for (int i = ; i <= p; ++i) printf("%d%c", ans[i], " \n"[i == p]);

     }

     return ;

 }

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

 #include <bits/stdc++.h>

 using namespace std;

 #define N 110

 int n, m, t;

 int g[N][N];

 int dp[N];

 int stk[N][N];

 int mx;

 map <string, int> mp;

 int get(string s)

 {

     if (mp.find(s) != mp.end()) return mp[s];

     else mp[s] = t++;

     return mp[s];

 }

 int DFS(int n, int ns, int dep)

 {

     if (ns == )

     {

         mx = max(mx, dep);

         return ;

     }

     int i, j, k, p, cnt;

     for (i = ; i < ns; ++i)

     {

         k = stk[dep][i];

         cnt = ;

         if (dep + n - k <= mx)

             return ;

         if (dep + dp[k] <= mx)

             return ;

         for (j = i + ; j < ns; ++j)

         {

             p = stk[dep][j];

             if (g[k][p])

                 stk[dep + ][cnt++] = p;

         }

         DFS(n, cnt, dep + );

     }

     return ;

 }

 int clique(int n)

 {

     int i, j, ns;

     for (mx = , i = n - ; i >= ; --i)

     {

         for (ns = , j = i + ; j < n; ++j)

         {

             if (g[i][j])

                 stk[][ns++] = j;

         }

         DFS(n, ns, );

         dp[i] = mx;

     }

     return mx;

 }

 vector <int> vec;

 void add()

 {

     vec.erase(unique(vec.begin(), vec.end()), vec.end());

     for (auto u : vec) for (auto v : vec)

         g[u][v] = g[v][u] = ;

     vec.clear();

 }

 int main()

 {

     while (scanf("%d%d", &n, &m) != EOF)

     {

         t = ;

         mp.clear();

         for (int i = ; i < m; ++i) for (int j = ; j < m; ++j) g[i][j] = ;

         []()

         {

             int op; char s[];

             for (int nn = ; nn <= n; ++nn)

             {

                 scanf("%d", &op);

                 if (op == )

                     add();

                 else

                 {

                     scanf("%s", s + );

                     vec.push_back(get(s + ));

                 }

             }

         }();

         add();

         for (int i = ; i < m; ++i) g[i][i] = ;

         //for (int i = 1; i <= m; ++i) for (int j = 1; j <= m; ++j) printf("%d %d %d\n", i, j, g[i][j]);

         printf("%d\n", clique(m));

     }

     return ;

 }

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$dp上去$

$vp的时候想到折半状压，但是当时是枚举子集来转移，复杂度大大增加..$

 #include <bits/stdc++.h>

 using namespace std;

 #define N 50

 #define M 1100010

 int n, m, t;

 int G[N][N];

 map <string, int> mp;

 int get(string s)

 {

     if (mp.find(s) != mp.end()) return mp[s];

     else mp[s] = ++t;

     return mp[s];

 }

 vector <int> vec;

 void add()

 {

     vec.erase(unique(vec.begin(), vec.end()), vec.end());

     for (auto u : vec) for (auto v : vec)

         G[u][v] = ;

     vec.clear();

 }

 int f[M], g[M];

 int main()

 {

     while (scanf("%d%d", &n, &m) != EOF)

     {

         t = ; mp.clear();

         memset(G, , sizeof G);

         int op; char s[N];

         for (int nn = ; nn <= n; ++nn)

         {

             scanf("%d", &op);

             if (op == ) add();

             else

             {

                 scanf("%s", s + );

                 vec.push_back(get(s + ));

             }

         }

         add();

         for (int i = ; i <= m; ++i)

             G[i][i] = ;

         int s1 = m / , s2 = m - s1;

         for (int i = ; i < ( << s1); i <<= ) f[i] = ;

         for (int i = ; i < ( << s2); i <<= ) g[i] = ;

         f[] = g[] = ;

         for (int i = ; i < ( << s1); ++i)

         {

             for (int j = ; j < s1; ++j) if (!((i >> j) & ))

             {

                 int flag = ;

                 for (int k = ; k < s1; ++k) if (((i >> k) & ) && G[k + ][j + ])

                 {

                     flag = ;

                     break;

                 }

                 f[i | ( << j)] = max(f[i | ( << j)], f[i] + flag);

             }

         }

         for (int i = ; i < ( << s2); ++i)

         {

             for (int j = ; j < s2; ++j) if (!((i >> j) & ))

             {

                 int flag = ;

                 for (int k = ; k < s2; ++k) if (((i >> k) & ) && G[s1 + k + ][s1 + j + ])

                 {

                     flag = ;

                     break;

                 }

                 g[i | ( << j)] = max(g[i | ( << j)], g[i] + flag);

             }

         }

         int res = ;

         for (int i = ; i < ( << s1); ++i)

         {

             int s3 = ( << s2) - ;

             for (int j = ; j < s1; ++j) if ((i >> j) & )

             {

                 for (int k = ; k < s2; ++k) if (G[j + ][s1 + k + ] && ((s3 >> k) & ))

                     s3 ^= ( << k);

             }

             res = max(res, f[i] + g[s3]);

         }

         printf("%d\n", res);

     }

     return ;

 }

法三：

为什么随机也行啊，能不能证明啊，喵喵喵..

 #include <bits/stdc++.h>

 using namespace std;

 #define ll long long

 #define N 50

 int n, m, t;

 int G[N][N], a[N];

 ll f[N];

 map <string, int> mp;

 int get(string s)

 {

     if (mp.find(s) != mp.end()) return mp[s];

     else mp[s] = ++t;

     return mp[s];

 }

 vector <int> vec;

 void add()

 {

     vec.erase(unique(vec.begin(), vec.end()), vec.end());

     for (auto u : vec) for (auto v : vec)

         G[u][v] = ;

     vec.clear();

 }

 int main()

 {

     while (scanf("%d%d", &n, &m) != EOF)

     {

         t = ; mp.clear();

         memset(G, , sizeof G);

         int op; char s[N];

         for (int nn = ; nn <= n; ++nn)

         {

             scanf("%d", &op);

             if (op == ) add();

             else

             {

                 scanf("%s", s + );

                 vec.push_back(get(s + ));

             }

         }

         add();

         for (int i = ; i <= m; ++i)

             G[i][i] = ;

         int res = ;

         for (int i = ; i <= m; ++i) a[i] = i;

         for (int i = ; i <= m; ++i)

         {

             f[i] = ;

             for (int j = ; j <= m; ++j) if (G[i][j])

                 f[i] |= (1ll << (j - ));

         }

         for (int t = ; t <= ; ++t)

         {

             random_shuffle(a + , a +  + m);

             ll g = ; int tmp = ;

             for (int i = ; i <= m; ++i) if (((g >> (a[i] - )) & 1ll) == )

             {

                 g |= f[a[i]];

                 ++tmp;

             }

             res = max(res, tmp);

         }

         printf("%d\n", res);

     }

     return ;

 }

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