[代码] bzoj 1500 维修数列(无旋treap)

Posted 2020-10-17 Anding

- 传送门 -

　http://www.lydsy.com/JudgeOnline/problem.php?id=1500
　

1500: [NOI2005]维修数列

Time Limit:?10 Sec??Memory Limit:?64 MB
Submit:?15301??Solved:?5063

Description

Input

输入的第1 行包含两个数N 和M(M ≤20 000)，N 表示初始时数列中数的个数，M表示要进行的操作数目。
第2行包含N个数字，描述初始时的数列。
以下M行，每行一条命令，格式参见问题描述中的表格。
任何时刻数列中最多含有500 000个数，数列中任何一个数字均在[-1 000, 1 000]内。
插入的数字总数不超过4 000 000个，输入文件大小不超过20MBytes。

Output

对于输入数据中的GET-SUM和MAX-SUM操作，向输出文件依次打印结果，每个答案（数字）占一行。

Sample Input

9 8
2 -6 3 5 1 -5 -3 6 3
GET-SUM 5 4
MAX-SUM
INSERT 8 3 -5 7 2
DELETE 12 1
MAKE-SAME 3 3 2
REVERSE 3 6
GET-SUM 5 4
MAX-SUM

Sample Output

-1
10
1
10

HINT

- 代码 -

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <ctime>
#include <cstdlib>
#define pii pair<int, int>
#define mp make_pair
#define ls C[rt][0]
#define rs C[rt][1]
using namespace std;

template <typename ty> void read(ty &x) {
  x = 0; int f = 1; char ch = getchar();
  while (ch > '9' || ch < '0') { if (ch == '-') f = -1; ch = getchar(); }
  while (ch >= '0' && ch <= '9') { x = x*10 + ch - '0'; ch = getchar(); }
  x *= f;
}
template <typename ty> ty Max(ty a, ty b) { return a > b ? a : b; }
template <typename ty> ty Min(ty a, ty b) { return a < b ? a : b; }
template <typename ty> int Chkmin(ty a, ty b) { return a > b ? a = b, 1 : 0; }
template <typename ty> int Chkmax(ty a, ty b) { return a < b ? a = b, 1 : 0; }

typedef long long LL;
typedef double db;

const int inf = 0x7fffffff;
const int N = 5e5 + 160;

int V[N], C[N][2], S[N], STK[N], KEY[N];
int TAG[N], REV[N], A[N], Q[4000016];
int LM[N], RM[N], TM[N], T[N];
int root, n, m, sz, tot;
int pos, t, c;

char OP[160];

void pushtag(int rt, int val) {

  if (!rt) return;
  V[rt] = val;
  T[rt] = S[rt] * val;
  LM[rt] = RM[rt] = TM[rt] = Max(val, val * S[rt]);
  TAG[rt] = 1;

}

void pushrev(int rt) {

  if (!rt) return;
  swap(C[rt][0], C[rt][1]);
  swap(LM[rt], RM[rt]);
  REV[rt] ^= 1;

}

void pushup(int rt) {

  if (!rt) return;
  S[rt] = S[ls] + S[rs] + 1;
  T[rt] = T[ls] + T[rs] + V[rt];
  LM[rt] = Max(LM[ls], T[ls] + V[rt] + Max(LM[rs], 0));
  RM[rt] = Max(RM[rs], T[rs] + V[rt] + Max(RM[ls], 0));
  TM[rt] = Max(TM[ls], Max(TM[rs], V[rt] + Max(LM[rs], 0) + Max(RM[ls], 0)));

}

void pushdown(int rt) {

  if (!rt) return;
 
  if (TAG[rt]) {
    
    pushtag(ls, V[rt]);
    pushtag(rs, V[rt]);
    TAG[rt] = REV[rt] = 0;

  }

  if (REV[rt]) {
    
    pushrev(ls);
    pushrev(rs);
    REV[rt] = 0;

  }

}

int new_node() {

  int a, b;
  read(a);
  if (tot) b = Q[tot--];
  else b = ++sz;
  V[b] = a; KEY[b] = rand();
  C[b][0] = C[b][1] = 0;
  S[b] = 1; TAG[b] = REV[b] = 0;
  LM[b] = RM[b] = TM[b] = T[b] = a;
  return b;

}

int build(int s) {

  int lst = 0, top = 0;
  for (int i = 1; i <= s; ++ i) {
    
    int tmp = new_node(); lst = 0;
    while (top && KEY[tmp] < KEY[STK[top]]) {
      pushup(STK[top]);
      lst = STK[top];
      STK[top--] = 0;
    }
    if (lst) C[tmp][0] = lst;
    if (top) C[STK[top]][1] = tmp;
    STK[++top] = tmp;

  }

  while (top) pushup(STK[top--]);
  return STK[1];

}

pii split(int rt, int k) {

  if (!rt) return mp(0, 0);
  
  pii tmp;
  pushdown(rt);
  
  if (k > S[C[rt][0]]) {
    tmp = split(C[rt][1], k - S[C[rt][0]] - 1);
    C[rt][1] = tmp.first; pushup(rt); tmp.first = rt;
  }
  else {
    tmp = split(C[rt][0], k);
    C[rt][0] = tmp.second; pushup(rt); tmp.second = rt;
  }
  
  return tmp;

}

int merge(int ra, int rb) {

  if (!ra) return rb;
  if (!rb) return ra;
  
  pushdown(ra);
  pushdown(rb);

  if (KEY[ra] < KEY[rb]) {
    C[ra][1] = merge(C[ra][1], rb);
    pushup(ra); return ra;
  }
  else {
    C[rb][0] = merge(ra, C[rb][0]);
    pushup(rb); return rb;
  }

}

void recycle(int rt) {

  if (!rt) return;
  Q[++tot] = rt;
  recycle(C[rt][0]);
  recycle(C[rt][1]);

}

void work1() {

  read(pos); read(t);
  pii a = split(root, pos);
  root = merge(a.first, merge(build(t), a.second));

}

void work2() {

  read(pos); read(t);
  pii a = split(root, pos - 1);
  pii b = split(a.second, t);
  recycle(b.first);
  root = merge(a.first, b.second);

}

void work3() {

  read(pos); read(t); read(c);
  if (!t) return;
  pii a = split(root, pos - 1);
  pii b = split(a.second, t);
  pushtag(b.first, c);
  root = merge(a.first, merge(b.first, b.second));

}

void work4() {

  read(pos); read(t);
  if (!t) return;
  pii a = split(root, pos - 1);
  pii b = split(a.second, t);
  pushrev(b.first);
  root = merge(a.first, merge(b.first, b.second));

}

void work5() {

  read(pos); read(t);
  if (!t) { printf("0\n"); return; }
  pii a = split(root, pos - 1);
  pii b = split(a.second, t);
  printf("%d\n", T[b.first]);
  root = merge(a.first, merge(b.first, b.second));

}

void work6() { printf("%d\n", TM[root]); }

int main () {

  read(n); read(m);

  LM[0] = RM[0] = TM[0] = V[0] = -inf;
  root = build(n);
  for (int i = 1; i <= m; ++ i) {
    scanf("%s", OP);
    if (OP[0] == 'I') work1();
    if (OP[0] == 'D') work2();
    if (OP[2] == 'K') work3();
    if (OP[0] == 'R') work4();
    if (OP[0] == 'G') work5();
    if (OP[2] == 'X') work6(); 
  }

  return 0;

}