Codeforces Round #532 (Div. 2) Solution

A. Roman and Browser

签到.

 #include <bits/stdc++.h>
 using namespace std;

 int n, k, a[];

 int get(int b)
 {
     int vis[];
     memset(vis, , sizeof vis);
     int c = b;
     while (c <= n)
     {
         vis[c] = ;
         c += k;
     }
     c = b - k;
     while (c >= )
     {
         vis[c] = ;
         c -= k;
     }
     int l = , r = ;
     for (int i = ; i <= n; ++i) if (!vis[i])
     {
         if (a[i] == ) ++l;
         else ++r;
     }
     return abs(l - r);
 }

 int main()
 {
     while (scanf("%d%d", &n, &k) != EOF)
     {
         for (int i = ; i <= n; ++i) scanf("%d", a + i);
         int res = ;
         for (int i = ; i <= n; ++i) res = max(res, get(i));
         printf("%d\n", res);
     }
     return ;
 }

B. Build a Contest

签到.

 #include <bits/stdc++.h>
 using namespace std;

 #define N 100010
 int n, m, a[N];
 int cnt[N]; 

 int main()
 {
     while (scanf("%d%d", &n, &m) != EOF)
     {
         for (int i = ; i <= m; ++i) scanf("%d", a + i);
         int need = n;
         memset(cnt, , sizeof cnt);
         for (int i = ; i <= m; ++i)
         {
             if (cnt[a[i]] == )
                 --need;
             ++cnt[a[i]];
             if (need) putchar('');
             else
             {
                 putchar('');
                 for (int j = ; j <= n; ++j)
                 {
                     --cnt[j];
                     if (!cnt[j])
                         ++need;
                 }
             }
         }
         puts("");
     }
     return ;
 }

C. NN and the Optical Illusion

签到.

 #include <bits/stdc++.h>
 using namespace std;

 const double eps = 1e-;
 const double PI = acos(-1.0);

 int main()
 {
     int n; double r;
     while (scanf("%d%lf", &n, &r) != EOF)
     {
         double ang1 = 2.0 * PI / n;
         double ang2 = (PI - ang1) / ;
         double R = (r * sin(ang1) * 1.0) / ( * sin(ang2) - sin(ang1));
         printf("%.10f\n", R);
     }
     return ;
 }

D. Dasha and Chess

Upsolved.

题意：

你有一个白王，对方有$666$个黑王

你每次可以八个方向走一步，对方可以任意移动一个黑王的位置

在$2000步以内如果你和某个黑王同行同列，你就赢了$

$给出你必胜的方案$

思路：

先走到中间，再考虑黑王个数最少的那个角落，向它的反方向走就可以了

考虑最差的情况，四个角很平均都有$166个左右$

那么也就是说另外三个角的个数加起来至少是$500个，而从中间走到某一角落只需要499步$

所以一定可以包围一个

 #include <bits/stdc++.h>
 using namespace std;

 #define N 1010
 #define pii pair <int, int>
 #define x first
 #define y second
 int G[N][N], x, y, cnt[];
 pii cor[N];

 void Move(int edx, int edy)
 {
     while (x != edx || y != edy)
     {
         if (x < edx && G[x + ][y] == ) ++x;
         else if (x > edx && G[x - ][y] == ) --x;
         if (y < edy && G[x][y + ] == ) ++y;
         else if (y > edy && G[x][y - ] == ) --y;
         printf("%d %d\n", x, y);
         fflush(stdout);
         int k, a, b;
         scanf("%d%d%d", &k, &a, &b);
         if (k == -) exit();
         G[cor[k].x][cor[k].y] = ;
         cor[k].x = a, cor[k].y = b;
         G[cor[k].x][cor[k].y] = ;
     }
 }

 int main()
 {
     while (scanf("%d%d", &x, &y) != EOF)
     {
         memset(G, , sizeof G);
         for (int i = ; i <= ; ++i)
         {
             scanf("%d%d", &cor[i].x, &cor[i].y);
             G[cor[i].x][cor[i].y] = ;
         }
         Move(, );
         memset(cnt, , sizeof cnt);
         for (int i = ; i <= ; ++i)
             for (int j = ; j <= ; ++j)
                 if (G[i][j])
                 {
                     if (i < x && j < y) ++cnt[];
                     else if (i < x && j > y) ++cnt[];
                     else if (i > x && j < y) ++cnt[];
                     else ++cnt[];
                 }
         if (cnt[] <= ) Move(, );
         else if (cnt[] <= ) Move(, );
         else if (cnt[] <= ) Move(, );
         else Move(, );
     }
     return ;
 }

E. Andrew and Taxi

Upsolved.

题意：

给出一张有向图，求改变一些边使得它没有环

改变一条边的花费是它的边权，要使得最大花费最小

输出最大花费和需要改变的边数

再输出相应的边

思路：

先二分答案，每次check的时候检查一下边权大于limit的边组成的图是否有环

如果有环，那么一定不行，如果没有环，那么一定可以

对于剩下的边，方向自定，拓扑序小的连向拓扑序大的

 #include <bits/stdc++.h>
 using namespace std;

 #define N 100010
 int n, m;
 struct Graph
 {
     struct node
     {
         int to, nx, w, id;
         node() {}
         node (int to, int nx, int w, int id) : to(to), nx(nx), w(w), id(id) {}
     }a[N << ];
     int head[N], pos;
     void init()
     {
         memset(head, , sizeof head);
         pos = ;
     }
     void add(int u, int v, int w, int id)
     {
         a[++pos] = node(v, head[u], w, id); head[u] = pos;
     }
 }G;
 #define erp(u) for (int it = G.head[u], v = G.a[it].to, w = G.a[it].w, id = G.a[it].id; it; it = G.a[it].nx, v = G.a[it].to, w = G.a[it].w, id = G.a[it].id)

 int degree[N], ord[N];
 bool check(int x)
 {
     memset(degree, , sizeof degree);
     for (int i = ; i <= n; ++i) erp(i) if (w > x)
         ++degree[v];
     int cnt = ;
     queue <int> q;
     for (int i = ; i <= n; ++i) if (!degree[i])
         q.push(i);
     while (!q.empty())
     {
         ++cnt;
         int u = q.front(); q.pop();
         ord[u] = cnt;
         erp(u) if (w > x && --degree[v] == )
             q.push(v);
     }
     return cnt >= n;
 }

 int main()
 {
     while (scanf("%d%d", &n, &m) != EOF)
     {
         G.init();
         for (int i = , u, v, w; i <= m; ++i)
         {
             scanf("%d%d%d", &u, &v, &w);
             G.add(u, v, w, i);
         }
         int l = , r = (int)1e9, res = -;
         while (r - l >= )
         {
             int mid = (l + r) >> ;
             if (check(mid))
             {
                 res = mid;
                 r = mid - ;
             }
             else
                 l = mid + ;
         }
         vector <int> vec;
         check(res);
         for (int i = ; i <= n; ++i) erp(i) if (w <= res && ord[i] > ord[v]) vec.push_back(id);
         printf("%d %d\n", res, (int)vec.size());
         for (int i = , len = vec.size(); i < len; ++i) printf("%d%c", vec[i], " \n"[i == len - ]);
     }
     return ;
 }

F. Ivan and Burgers

Upsolved.

题意：

询问区间任意数异或最大值

思路：

维护前缀线性基，但是要注意将$位置id大的数更换基底$

$因为这样这些基底被用户更新答案的概率就更高$

 #include <bits/stdc++.h>
 using namespace std;

 #define N 500010
 int n, q, c[N];
 int pos[N][], p[N][];

 void work(int x)
 {
     for (int i = ; i <= ; ++i)
     {
         pos[x][i] = pos[x - ][i];
         p[x][i] = p[x - ][i];
     }
     int tmp = c[x];
     int id = x;
     for (int i = ; i >= ; --i)
     {
         if ((tmp >> i) & )
         {
             if (!p[x][i])
             {
                 p[x][i] = tmp;
                 pos[x][i] = id;
                 return;
             }
             if (pos[x][i] < id)
             {
                 swap(tmp, p[x][i]);
                 swap(id, pos[x][i]);
             }
             tmp ^= p[x][i];
         }
     }
 }

 int ask(int l, int r)
 {
     int res = ;
     for (int i = ; i >= ; --i) if (pos[r][i] >= l && (res ^ p[r][i]) > res)
         res ^= p[r][i];
     return res;
 }

 int main()
 {
     while (scanf("%d", &n) != EOF)
     {
         for (int i = ; i <= n; ++i) scanf("%d", c + i);
         for (int i = ; i < ; ++i) pos[][i] = p[][i] = ;
         for (int i = ; i <= n; ++i) work(i);
         scanf("%d", &q);
         for (int qq = , l, r; qq <= q; ++qq)
         {
             scanf("%d%d", &l, &r);
             printf("%d\n", ask(l, r));
         }

     }
     return ;
 }