[SHOI2015]脑洞治疗仪

嘟嘟嘟




这题其实就是一个线段树维护最大连续和的水题。

别的操作不说，操作1只要二分找区间前\(k\)个0即可。

需要注意的是，因为操作1两区间可能有交，因此要先清空再二分查询……

复杂度\(O(n log ^ 2 n)\)。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2e5 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m;
struct Tree
{
  int l, r, sum, lzy;
  int lmax, rmax, imax;
  In Tree operator + (const Tree& oth)const
  {
    Tree ret;
    ret.l = l, ret.r = oth.r;
    ret.sum = sum + oth.sum, ret.lzy = -1;
    ret.lmax = lmax;
    if(ret.lmax == r - l + 1) ret.lmax += oth.lmax;
    ret.rmax = oth.rmax;
    if(ret.rmax == oth.r - oth.l + 1) ret.rmax += rmax;
    ret.imax = max(max(imax, oth.imax), rmax + oth.lmax);
    return ret;
  }
}t[maxn << 2];
In void build(int L, int R, int now)
{
  t[now].l = L, t[now].r = R;
  t[now].lzy = -1;
  if(L == R) {t[now].sum = 1;return;}
  int mid = (L + R) >> 1;
  build(L, mid, now << 1);
  build(mid + 1, R, now << 1 | 1);
  t[now] = t[now << 1] + t[now << 1 | 1];
}
In void change(int now, int d)
{
  t[now].sum = (t[now].r - t[now].l + 1) * d;
  t[now].lzy = d;
  t[now].lmax = t[now].rmax = t[now].imax = (t[now].r - t[now].l + 1) * (d ^ 1);
}
In void pushdown(int now)
{
  if(~t[now].lzy)
    {
      change(now << 1, t[now].lzy), change(now << 1 | 1, t[now].lzy);
      t[now].lzy = -1;
    }
}
In void update(int L, int R, int now, int d)
{
  if(t[now].l == L && t[now].r == R) {change(now, d); return;}
  pushdown(now);
  int mid = (t[now].l + t[now].r) >> 1;
  if(R <= mid) update(L, R, now << 1, d);
  else if(L > mid) update(L, R, now << 1 | 1, d);
  else update(L, mid, now << 1, d), update(mid + 1, R, now << 1 | 1, d);
  t[now] = t[now << 1] + t[now << 1 | 1];
}
In int query_sum(int L, int R, int now)
{
  if(t[now].l == L && t[now].r == R) return t[now].sum;
  pushdown(now);
  int mid = (t[now].l + t[now].r) >> 1;
  if(R <= mid) return query_sum(L, R, now << 1);
  else if(L > mid) return query_sum(L, R, now << 1 | 1);
  else return query_sum(L, mid, now << 1) + query_sum(mid + 1, R, now << 1 | 1);
}
In Tree query_con(int L, int R, int now)
{
  if(t[now].l == L && t[now].r == R) return t[now];
  pushdown(now);
  int mid = (t[now].l + t[now].r) >> 1;
  if(R <= mid) return query_con(L, R, now << 1);
  else if(L > mid) return query_con(L, R, now << 1 | 1);
  else return query_con(L, mid, now << 1) + query_con(mid + 1, R, now << 1 | 1);
}

int main()
{
  n = read(), m = read();
  build(1, n, 1);
  for(int i = 1; i <= m; ++i)
    {
      int op = read(), l1 = read(), r1 = read();
      if(op == 0) update(l1, r1, 1, 0);
      else if(op == 2) write(query_con(l1, r1, 1).imax), enter;
      else
	{
	  int l2 = read(), r2 = read();
	  int sum = query_sum(l1, r1, 1);
	  update(l1, r1, 1, 0);
	  if(!sum) continue;
	  int L = l2, R = r2;
	  while(L < R)
	    {
	      int mid = (L + R) >> 1;
	      if(mid - l2 + 1 - query_sum(l2, mid, 1) >= sum) R = mid;
	      else L = mid + 1;
	    }
	  update(l2, L, 1, 1);
	}
    }
  return 0;
}