CF1097E Egor and an RPG game

最少反链划分数 = 最长链。实现：每次找出所有极大元作为一个反链。

任意长度小于k * (k + 1) / 2的排列都能被划分为不多于k个单调序列。且这是一个紧的上界。

然后这题就可以切了。

题意：给定长为n的排列，设任长为n的排列都能被划分为不多于k个单调序列，把给定排列划分为不多于k个单调序列。

解：先求出k，然后求LIS，如果LIS >= k那么扔掉LIS递归。否则划分成反链即可。用链表实现。

 #include <bits/stdc++.h>

 const int N = , INF = 0x3f3f3f3f;

 int p[N], cnt;
 std::vector<int> v[N];

 struct Node {
     int nex, pre;
 }node[N]; int tp, head = N - , tail = N - ;

 int stk[N], top, f[N], rest, stk2[N], fr[N];
 bool vis[N];

 inline void del(int x) {
     int nex = node[x].nex, pre = node[x].pre;
     node[nex].pre = pre;
     node[pre].nex = nex;
     return;
 }

 inline int getLIS() {
     top = ;
     for(int i = node[head].nex; i != tail; i = node[i].nex) {
         int l = , r = top;
         while(l < r) {
             int mid = (l + r + ) >> ;
             if(stk[mid] < p[i]) {
                 l = mid;
             }
             else {
                 r = mid - ;
             }
         }
         f[i] = r + ;
         fr[i] = stk2[r];
         stk[r + ] = p[i];
         stk2[r + ] = i;
         if(r == top) ++top;
     }
     return top;
 }

 inline void erase() {
     int x = stk2[top];
     ++cnt;
     while(x) {
         del(x);
         v[cnt].push_back(p[x]);
         x = fr[x];
     }
     std::reverse(v[cnt].begin(), v[cnt].end());
     return;
 }

 inline void split() {

     while(rest) {
         ++cnt;
         int last = -;
         for(int i = node[tail].pre; i != head; i = node[i].pre) {
             if(p[i] > last) {
                 v[cnt].push_back(i);
                 last = p[i];
                 //printf("inside %d \n", i);
                 rest--;
             }
         }
         std::reverse(v[cnt].begin(), v[cnt].end());
         int len = v[cnt].size();
         for(int i = ; i < len; i++) {
             del(v[cnt][i]);
             //printf("now : %d \n", v[cnt][i]);
             v[cnt][i] = p[v[cnt][i]];
             //printf("now : %d \n", v[cnt][i]);
         }
     }

     return;
 }

 inline void solve() {
     int n;
     scanf("%d", &n);
     for(int i = ; i <= n; i++) {
         scanf("%d", &p[i]);
         node[i].nex = ;
         if(i > ) {
             node[i].pre = i - ;
             node[i - ].nex = i;
         }
         else {
             node[i].pre = head;
             node[head].nex = i;
         }
     }
     node[n].nex = tail;
     node[tail].pre = n;
     cnt = ;
     rest = n;
     int K = ;
     while(K * (K + ) /  <= n) {
         ++K;
     }
     //printf("K = %d \n", K);
     /// use K-1
     while(rest) {
         int len = getLIS();
         //printf("len = %d \n", len);
         if(len >= K) {
             erase();
             K--;
             rest -= len;
         }
         else {
             split();
             rest = ;
         }
     }
     printf("%d \n", cnt);
     for(int i = ; i <= cnt; i++) {
         int len = v[i].size();
         printf("%d ", len);
         for(int j = ; j < len; j++) {
             printf("%d ", v[i][j]);
         }
         puts("");
         v[i].clear();
     }
     tp = ;
     return;
 }

 int main() {

     int T;
     scanf("%d", &T);
     while(T--) {
         solve();
     }

     return ;
 }

AC代码