CF1303F - Number of Components

Solution

思路还是有点妙的。

容易想到并查集，但是并查集不容易维护删边，怎么办呢？

我们考虑拆贡献，把加边的贡献和删边的贡献拆开，分别维护。
只加边就是四连通加边，算一下新增多少个连通块。
只删边的贡献可以从后往前做，变成加边，对答案的贡献就是新增连通块个数的相反数（因为删边就是加边的逆过程，贡献相反）。

并查集维护即可。

时间复杂度O(qlog⁡n)O(q\log n)O(qlogn)。

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 = 100005;
const int MAXM = 2000005;
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(); !isdigit(c) && c != EOF; c = gc());}inline void reads(char *st) {char c;int n = 0;getc(st[++ n]);for (c = gc(); isdigit(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;vector<PR> A[MAXM], _A[MAXM];
int dx[4] = {0, 0, -1, 1};
int dy[4] = {1, -1, 0, 0};
int f[MAXN], col[305][305], Ans[MAXM], _Ans[MAXM], n, m, Case;
int getid(int x, int y) { return (x - 1) * m + y; }
int find(int x) { return f[x] == x ? f[x] : f[x] = find(f[x]); }
void solve(vector<PR> &A, int opt) {for (auto v : A) f[v.fi] = v.fi;for (auto v : A) {int x = (v.fi - 1) / m + 1, y = (v.fi - 1) % m + 1, t = 1;col[x][y] = 1;for (int i = 0; i < 4 ; ++ i) {int _x = x + dx[i], _y = y + dy[i], p = getid(_x, _y);if (_x < 1 || _y < 1 || _x > n || _y > m || col[x][y] != col[_x][_y] || find(p) == find(v.fi)) continue; f[find(p)] = find(v.fi);-- t;}if (opt == 1) Ans[v.se] += t;else Ans[v.se] -= t;}   for (auto v : A) col[(v.fi - 1) / m + 1][(v.fi - 1) % m + 1] = 0;
}
signed main() {
#ifndef ONLINE_JUDGEfreopen("a.in", "r", stdin);
#endifread(n), read(m), read(Case);int MX = max(1000, 2000000 / n / m + 1);for (int i = 1; i <= n * m ; ++ i) A[0].PB(MP(i, 0));for (int i = 1; i <= Case ; ++ i) {int x, y, z;read(x), read(y), read(z);_A[col[x][y]].PB(MP(getid(x, y), i));col[x][y] = z;A[col[x][y]].PB(MP(getid(x, y), i));   }for (int i = 1; i <= n * m ; ++ i) _A[col[(i - 1) / m + 1][(i - 1) % m + 1]].PB(MP(i, Case + 1));for (int i = 0; i <= MX ; ++ i) reverse(_A[i].begin(), _A[i].end());for (int i = 0; i <= MX ; ++ i) solve(A[i], 1);for (int i = 0; i <= MX ; ++ i) solve(_A[i], -1);for (int i = 1; i <= Case ; ++ i) Ans[i] += Ans[i - 1];for (int i = 1; i <= Case; ++ i) print(Ans[i]), putc('\n');return 0;
}

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CF1303F - Number of Components(并查集)

CF1303F - Number of Components

Solution

Code

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