CF1386C. Joker

Solution

难得有一道可以整体二分的题。

有一个基本的思路是：考虑求出$An{s}_i$表示最小的右端点，满足$[1,i]\cup [An{s}_i,m]$中存在奇环。

考虑到$An{s}_i$序列是非减的，因此我们可以$twopointers+LCT$维护生成树以及是否存在奇环，时间复杂度$O(n\log n)$。

不过还有一种$O(n{\log }^{2}n)$的整体二分做法：
令$Solve(l,r,L,R)$表示求出$i\in [l,r],An{s}_i\in [L,R]$的所有$An{s}_i$。有一个精妙的想法是：我们去保证$Solve(l,r,L,R)$时$[1,l]\cup [R,n]$的边已经加入 ，那么$i=mid$的答案就可以通过加入$[l+1,mid]$以及不超过$O(R-L)$条边求出，然后分治下去。

总时间复杂度：$O(n{\log }^{2}n)$。

Code

#include <bits/stdc++.h>using namespace std;template<typename T> inline bool upmin(T &x, T y) { return y < x ? x = y, 1 : 0; }
template<typename T> inline bool upmax(T &x, T y) { return x < y ? x = y, 1 : 0; }#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondtypedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int, int> PR;
typedef vector<int> VI; const lod eps = 1e-9;
const lod pi = acos(-1);
const int oo = 1 << 30;
const ll loo = 1ll << 60;
const int mods = 1e9 + 7;
const int inv2 = (mods + 1) >> 1;
const int MAXN = 200005;
const int INF = 0x3f3f3f3f; //1061109567
/*--------------------------------------------------------------------*/namespace FastIO{constexpr int SIZE = (1 << 21) + 1;int num = 0, f;char ibuf[SIZE], obuf[SIZE], que[65], *iS, *iT, *oS = obuf, *oT = obuf + SIZE - 1, c;#define gc() (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)), (iS == iT ? EOF : *iS ++)) : *iS ++)inline void flush() {fwrite(obuf, 1, oS - obuf, stdout);oS = obuf;}inline void putc(char c) {*oS ++ = c;if (oS == oT) flush();}inline void getc(char &c) {for (c = gc(); c != '<' && c != '>' && c != EOF; c = gc());}inline void reads(char *st) {char c;int n = 0;getc(st[++ n]);for (c = gc(); c == '<' || c == '>' ; c = gc()) st[++ n] = c;st[n + 1] = '\0';}template<class I>inline void read(I &x) {for (f = 1, c = gc(); c < '0' || c > '9' ; c = gc()) if (c == '-') f = -1;for (x = 0; c >= '0' && c <= '9' ; c = gc()) x = (x << 3) + (x << 1) + (c & 15);x *= f;}template<class I>inline void print(I x) {if (x < 0) putc('-'), x = -x;if (!x) putc('0');while (x) que[++ num] = x % 10 + 48, x /= 10;while (num) putc(que[num --]);}struct Flusher_{~Flusher_(){flush();}} io_Flusher_;
}
using FastIO :: read;
using FastIO :: putc;
using FastIO :: reads;
using FastIO :: print;PR E[MAXN];
int f[MAXN], col[MAXN], Ans[MAXN], num[MAXN], top = 0, n, m, Case;
struct stknode{ int x, p, y, q; } stk[MAXN];int find(int x) { return f[x] == x ? f[x] : find(f[x]); }
int getcol(int x) { return f[x] == x ? col[x] : getcol(f[x]) ^ col[x]; }
int Add(int u, int v) {int x = find(u), y = find(v);if (num[x] > num[y]) swap(x, y);if (x != y) {stk[++ top] = (stknode){x, col[x], y, num[y]};col[x] ^= getcol(u) ^ getcol(v) ^ 1;f[x] = y, num[y] += num[x];return 0;}return (getcol(u) == getcol(v));
}
void del(int x) {while (top > x) {f[stk[top].x] = stk[top].x;col[stk[top].x] = stk[top].p;num[stk[top].y] = stk[top].q;-- top;}
}
void solve(int l, int r, int L, int R, int flag) {
//	cout << l << " " << r << " " << L << " " << R << " " << flag << endl;if (l > r) return;if (flag || L == R) {for (int i = l; i <= r ; ++ i) Ans[i] = R;return;}int mid = (l + r) >> 1, lst = top, Flag = 0;for (int i = l + 1; i <= mid ; ++ i) Flag |= Add(E[i].fi, E[i].se);if (Flag) Ans[mid] = R;else {int Lst = top;for (int i = R - 1; i >= L ; -- i)if (Add(E[i].fi, E[i].se)) { Ans[mid] = i; break; }del(Lst);}Flag |= Add(E[mid + 1].fi, E[mid + 1].se);solve(mid + 1, r, Ans[mid], R, Flag);del(lst);Flag = 0;for (int i = R - 1; i >= Ans[mid] ; -- i) Flag |= Add(E[i].fi, E[i].se);solve(l, mid - 1, L, Ans[mid], Flag);
}
signed main() {
#ifndef ONLINE_JUDGEfreopen("a.in", "r", stdin);
#endifread(n), read(m), read(Case);for (int i = 1; i <= n ; ++ i) f[i] = i, num[i] = 1;for (int i = 1; i <= m ; ++ i) read(E[i].fi), read(E[i].se);solve(0, m - 1, 1, m + 1, 0);
//	for (int i = 1; i <= n ; ++ i) cout << Ans[i] << endl;while (Case --) {int l, r;read(l), read(r);if (Ans[l - 1] > r) putc('Y'), putc('E'), putc('S');else putc('N'), putc('O');putc('\n');}return 0;
}