[POI2008]KLO && POC

题意：给定一个序列 s1, s2,...sn,以及一个k，求一个连续的k个数，把s[i]...s[i+k-1]变成一个数s'，使得sigma(|s[j]-s'|)(i<=j<=i+k-1)最小

思路：最近无聊切起poi。。顿感智商不够。。

这道题很容易想到把这些数变成中位数最优。。那么就可以用平衡树维护了。。

当然，直接用set维护也就够了。。不过为了练练代码。。我还是写了下splay。。

code:

 /*
  * Author:  Yzcstc
  * Created Time:  2014/11/8 19:03:57
  * File Name: klo.cpp
  */
 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<map>
 #include<set>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<ctime>
 #define repf(i, a, b) for (int i = (a); i <= (b); ++i)
 #define M0(x)  memset(x, 0, sizeof(x))
 #define Inf  0x7fffffff
 using namespace std;
 const int maxn = ;
 typedef long long ll;
 int n, k;
 int a[];
 #define L ch[x][0]
 #define R ch[x][1]
 #define KT ch[ch[root][0]][1]
 struct SplayTree{
       int sz[maxn], pre[maxn], ch[maxn][], v[maxn];
       ll sum[maxn];
       int m, root;
       void push_up(const int& x){
            sz[x] = sz[L] + sz[R] + ;
            sum[x] = sum[L] + sum[R] + v[x];
       }
       void init(){
            M0(sz), M0(pre), M0(ch), M0(ch), M0(v);
       }
       void push_down(const int& x){
       }
       void rotate(int &x, const int f){
             int y = pre[x], z = pre[y];
             push_down(y), push_down(x);
             ch[y][!f] = ch[x][f];
             pre[ch[x][f]] = y;
             pre[x] = pre[y];
             if (z) ch[z][ch[z][] == y] = x;
             ch[x][f] = y;
             pre[y] = x;
             push_up(y);
       }
       void splay(int &x, const int g){
             push_down(x);
             while (pre[x] != g){
                 int y = pre[x], z = pre[y];
                 if (z == g) rotate(x, ch[y][] == x);
                 else {
                      int f = (ch[z][] == y);
                      ch[y][!f] ? rotate(y, f) : rotate(x, !f);
                      rotate(x, f);
                 }
             }
             push_up(x);
             if (!g) root = x;
       }
       void rto(int k,const int g, int& y){
             int x = root;
             while (sz[L] +  != k){
                   push_down(x);
                   if (sz[L] >= k) x = L;
                   else {
                        k -= sz[L] + ;
                        x = R;
                   }
             }
             y = x;
             splay(x, g);
       }
       int search(int x, int val){
             int y = -;
             while (x){
                  y = x;
                  if (val <= v[x]) y = x, x = L;
                  else y = x, x = R;
             }
             return y;
       }
       void insert(int x, int val){
            int y = search(root, val);
            ch[y][val > v[y]] = x;
            pre[x] = y;
            while (y) push_up(y), y = pre[y];
            splay(x, );
       }
       int findpre(int x){
            x = L;
            while (R) x = R;
            return x;
       }
       void split(int x){
            splay(x, );
            int lt = findpre(x);
            if (lt){
                   splay(lt, x);
                   root = lt, pre[root] = ;
                   ch[root][] = R;
                   if (R) pre[R] = root;
            } else root = R, pre[R] = ;
            push_up(root);
            L = R = ;
       }
       ll query(const int& n, int &vv){
             int k = (n+) / , x;
             if (!(n & )) ++k;
             rto(k, , x);
             ll res = (ll)v[x] * (k-) - sum[ch[root][]] + sum[ch[root][]] - (ll)v[x] * (n-k);
             vv = v[x];
             return res;
       }
 } S;

 void init(){
      for (int i = ; i <= n; ++i)
             scanf("%d", &a[i]);
 }

 void solve(){
      if (k == ){
            puts("");
            for (int i = ; i <= n; ++i)
                printf("%d\n", a[i]);
            return;
      }
      S.root = , S.sz[] = , S.sum[] = S.v[] = a[];
      for (int i = ; i <= n; ++i)
            S.sz[i] = , S.v[i] = S.sum[i] = a[i];
      for (int i = ; i <= k; ++i)
            S.insert(i, a[i]);
      int ans_pos = , ans_val = , val;
      ll ans = S.query(k, ans_val), tmp;
      for (int i = k + ; i <= n; ++i){
             S.split(i-k);
             S.insert(i, a[i]);
             tmp = S.query(k, val);
             if (tmp < ans)
                  ans = tmp, ans_val = val, ans_pos = i - k + ;
      }
      cout << ans << endl;
      for (int i = ans_pos; i <= ans_pos+k-; ++i) a[i] = ans_val;
      for (int i = ; i <= n; ++i) printf("%d\n", a[i]);
 }

 int main(){
 //    freopen("a.in", "r", stdin);
 //    freopen("a.out", "w", stdout);
     while (scanf("%d%d", &n, &k) != EOF){
           init();
           solve();
     }
     return ;
 }

题意：有 n(n<=1000)个串，每个串的程度都<=100，有 m 次操作，每次操作都是将第 p1个串的第 w1 个字母和第 p2 个串的第 w2 个字母交换，问对于每一个串 i，在这些操作执行的过程中，它最多几个串相同过。

思路：由于每个串只有100，我们可以用n*m的时间进行hash。。

那么剩下来就是直接维护hash值为某一个数的集合了。。。平衡树可维护。。

code：

 /*
  * Author:  Yzcstc
  * Created Time:  2014/11/8 13:53:39
  * File Name: poc.cpp
  */
 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<map>
 #include<set>
 #include<vector>
 #include<queue>
 #include<stack>
 #include<ctime>
 #define M0(x)  memset(x, 0, sizeof(x))
 #define Inf  0x7fffffff
 using namespace std;
 #define M 10003
 const int maxn = ;
 /*** splay-tree***/
 #define L ch[x][0]
 #define R ch[x][1]
 int sz[maxn], pre[maxn], rt[maxn], ms[maxn], ch[maxn][], lz[maxn];

 void Splay_init(){
      M0(sz), M0(pre), M0(rt), M0(ms),  M0(ch), M0(lz);
 }

 void pushdown(const int &x){
      if (lz[x]){
           if (L) lz[L] = max(lz[L], lz[x]), ms[L] = max(ms[L], lz[x]);
           if (R) lz[R] = max(lz[R], lz[x]), ms[R] = max(ms[R], lz[x]);
           lz[x] = ;
      }
 }

 void pushup(int x){}

 void rotate(int &x, const int f){
      int y = pre[x], z = pre[y];
      pushdown(y), pushdown(x);
      ch[y][!f] = ch[x][f];
      pre[ch[x][f]] = y;
      pre[x] =  pre[y];
      if (z) ch[z][ch[z][] == y] = x;
      ch[x][f] = y;
      pre[y] = x;
      pushup(y);
 }

 void splay(int &x, const int g, int &root){
      pushdown(x);
      while (pre[x] != g){
           int y = pre[x],  z = pre[y];
           if (z == g) rotate(x, ch[y][] == x);
           else {
                int f = (ch[z][] == y);
                ch[y][!f] ? rotate(y, f) : rotate(x, !f);
                rotate(x, f);
           }
      }
      pushup(x);
      if (!g) root = x;
 }

 void update(int &root, int s){
      lz[root] = max(lz[root], s);
      ms[root] = max(ms[root], s);
 }

 int right(int x){
      while (R) x = R;
      return x;
 }

 void insert(int& root, int x, int size){
      int p = right(root);
      splay(p, , root);
      ch[p][] = x, pre[x] = p;
      update(root, size);
 }

 void split(int& root, int x){
      splay(x, , root);
      int lt = ch[x][];
      lt = right(lt);
      if (lt){
           splay(lt, x, root), root = lt;
           if (R) pre[R] = root;
           pre[root] = , ch[root][] = R;
      } else
           root = R, pre[R] = ;
      L = R = ;
 }

 /*** splay-end***/
 map<int, int> mp;
 char s[][];
 int hs[], n, m, q, pw[];

 inline int Hash(const char s[]){
     int res = ;
     for (int i = ; i < m; ++i)
          res = res * M + s[i];
     return res;
 }

 void init(){
      pw[m-] = ;
      for (int i = m-; i >= ; --i)
            pw[i] = pw[i+] * M;
      for (int i = ; i <= n; ++i){
           scanf("%s", s[i]);
           hs[i] = Hash(s[i]);
      }
 }

 int f[maxn];
 void solve(){
      Splay_init();
      mp.clear();
      int tot = , u, v;
      for (int i = ; i <= n; ++i){
            u = hs[i], v = mp[u];
            if (v)
                   insert(rt[v], i, ++sz[v]);
            else {
                   v = mp[u] = ++tot;
                   sz[v] = ; rt[v] = i;
                   update(rt[v], );
            }
      }
      int a, x1, b, x2;
      int h1, h2;
      while (q--){
            scanf("%d%d%d%d", &a, &x1, &b, &x2);
            x1--, x2--;
            if (a == b){
                    h1 = hs[a] + pw[x1] * (s[b][x2] - s[a][x1]) + pw[x2] * (s[a][x1] - s[b][x2]);
                    u = mp[hs[a]], split(rt[u], a);
                    if (--sz[u] == ) mp[hs[a]] = rt[u] = ;
                    v = mp[h1];
                    if (v)
                           insert(rt[v], a,  ++sz[v]);
                    else {
                           v = mp[h1] = ++tot;
                           sz[v] = ; rt[v] = a;
                           update(rt[v], );
                    }
                    hs[a] = h1;
                    swap(s[a][x1], s[b][x2]);
                    continue;
            }
            h1 = hs[a] + pw[x1] * (s[b][x2] - s[a][x1]);
            h2 = hs[b] + pw[x2] * (s[a][x1] - s[b][x2]);
            u = mp[hs[a]], split(rt[u], a);
            if (--sz[u] == ) mp[hs[a]] = rt[u] = ;
            u = mp[hs[b]], split(rt[u], b);
            if (--sz[u] == ) mp[hs[b]]= rt[u] = ;
            v = mp[h1];
            if (v)
                   insert(rt[v], a, ++sz[v]);
            else {
                   v = mp[h1] = ++tot;
                   sz[v] = ; rt[v] = a;
                   update(rt[v], );
            }
            v = mp[h2];
            if (v)
                   insert(rt[v], b, ++sz[v]);
            else {
                   v = mp[h2] = ++tot;
                   sz[v] = ; rt[v] = b;
                   update(rt[v], );
            }
            hs[a] = h1, hs[b] = h2;
            swap(s[a][x1], s[b][x2]);
      }
      for (int i = ; i <= n; ++i){
           u = mp[hs[i]];
           splay(i, , rt[u]);
           f[i] = ms[i];
      }
      for (int i = ; i <= n; ++i)
             printf("%d\n", f[i]);
 }

 int main(){
 //    freopen("a.in", "r", stdin);
 //    freopen("a.out", "w", stdout);
     while (scanf("%d%d%d", &n, &m, &q) != EOF){
           init();
           solve();
     }
     fclose(stdin); fclose(stdout);
     return ;
 }