Codeforces Round #539 (Div. 1) 1109F. Sasha and Algorithm of Silence's Sounds LCT+线段树 (two pointers)

题解请看 Felix-Lee的CSDN博客

写的很好，不过最后不用判断最小值是不是1，因为[i,i]只有一个点，一定满足条件，最小值一定是1。

CODE

写完就A，刺激。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define X first
#define Y second
inline void read(int &x) {
    int flag = 1; char ch;
    while(!isdigit(ch=getchar()))if(ch=='-')flag=-flag;
    for(x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());
    x*=flag;
}
const int MAXN = 1005;
const int MAXP = 1000005;
const int INF = 1e9;
namespace LCT {
	#define ls ch[x][0]
	#define rs ch[x][1]
	int ch[MAXP][2], fa[MAXP];
	bool rev[MAXP];
	inline bool isr(int x) { return ch[fa[x]][0] != x && ch[fa[x]][1] != x; }
	inline bool get(int x) { return ch[fa[x]][1] == x; }
	inline void mt(int x) { if(rev[x]) rev[x]^=1, rev[ls]^=1, rev[rs]^=1, swap(ls, rs); }
	void mtpath(int x) { if(!isr(x)) mtpath(fa[x]); mt(x); }
	inline void rot(int x) {
		int y = fa[x], z = fa[y]; bool l = get(x), r = l^1;
		if(!isr(y)) ch[z][get(y)] = x;
		fa[ch[x][r]] = y; fa[y] = x; fa[x] = z;
		ch[y][l] = ch[x][r]; ch[x][r] = y;
	}
	inline void splay(int x) {
		mtpath(x);
		for(; !isr(x); rot(x))
			if(!isr(fa[x])) rot(get(fa[x]) == get(x) ? fa[x] : x);
	}
	inline void access(int x) { int y = 0;
		for(; x; x = fa[y=x]) splay(x), ch[x][1] = y;
	}
	inline void bert(int x) { access(x), splay(x), rev[x]^=1; }
	inline int sert(int x) { access(x), splay(x); for(; ch[x][0]; x=ch[x][0]); return x; }
	inline void link(int x, int y) { bert(x); if(sert(y) != x) fa[x] = y; }
	inline void cut(int x, int y) { bert(x); sert(y); fa[x] = ch[y][0] = 0; }
	inline bool connect(int x, int y) { bert(x); return sert(y) == x; }
}
using namespace LCT;
int n, m, a[MAXN][MAXN], tot;
#define pii pair<int, int>
pii pos[MAXP], val[MAXP<<2];
int lz[MAXP<<2];
inline pii merge(pii A, pii B) {
	if(A.X < B.X) return A;
	else if(A.X > B.X) return B;
	return pii(A.X, A.Y + B.Y);
}
inline void upd(int i) { val[i] = merge(val[i<<1], val[i<<1|1]); }
inline void pd(int i) {
	if(lz[i]) {
		val[i<<1].X += lz[i], lz[i<<1] += lz[i];
		val[i<<1|1].X += lz[i], lz[i<<1|1] += lz[i];
		lz[i] = 0;
	}
}
void build(int i, int l, int r) {
	if(l == r) { val[i] = pii(0, 1); return; }
	int mid = (l + r) >> 1;
	build(i<<1, l, mid);
	build(i<<1|1, mid+1, r);
	upd(i);
}
void modify(int i, int l, int r, int x, int y, int v) {
	if(x <= l && r <= y) { val[i].X += v, lz[i] += v; return; }
	pd(i);
	int mid = (l + r) >> 1;
	if(x <= mid) modify(i<<1, l, mid, x, y, v);
	if(y > mid) modify(i<<1|1, mid+1, r, x, y, v);
	upd(i);
}
pii query(int i, int l, int r, int x, int y) {
	if(x <= l && r <= y) return val[i];
	pd(i);
	int mid = (l + r) >> 1;
	pii re = pii(INF, 0);
	if(x <= mid) re = merge(re, query(i<<1, l, mid, x, y));
	if(y > mid) re = merge(re, query(i<<1|1, mid+1, r, x, y));
	return re;
}
const int dx[4] = { 0, 0, 1, -1 };
const int dy[4] = { 1, -1, 0, 0 };
inline bool chkout(int x, int y) {
	return x < 1 || y < 1 || x > n || y > m;
}
inline bool check(int l, int r) {
	for(int i = 0; i < 3; ++i) {
		int x = pos[r].X + dx[i];
		int y = pos[r].Y + dy[i];
		if(chkout(x, y) || a[x][y] < l || a[x][y] > r) continue;
		for(int j = i+1; j < 4; ++j) {
			int u = pos[r].X + dx[j];
			int v = pos[r].Y + dy[j];
			if(chkout(u, v) || a[u][v] < l || a[u][v] > r) continue;
			if(connect(a[x][y], a[u][v])) return 0;
		}
	}
	return 1;
}
inline void del(int l) {
	for(int i = 0; i < 4; ++i) {
		int x = pos[l].X + dx[i];
		int y = pos[l].Y + dy[i];
		if(chkout(x, y) || !connect(l, a[x][y])) continue;
		cut(l, a[x][y]);
	}
}
inline void solve(int l, int r) {
	for(int i = 0; i < 4; ++i) {
		int x = pos[r].X + dx[i];
		int y = pos[r].Y + dy[i];
		if(chkout(x, y) || a[x][y] < l || a[x][y] > r) continue;
		link(a[x][y], r);
		modify(1, 1, tot, 1, a[x][y], -1);
	}
	modify(1, 1, tot, l, r, 1);
}
int main () {
	scanf("%d%d", &n, &m); tot = n*m;
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j)
			scanf("%d", &a[i][j]), pos[a[i][j]] = pii(i, j);
	build(1, 1, tot);
	LL ans = 0;
	for(int i = 1, j = 1; i <= tot; ++i) {
		while(!check(j, i)) del(j++);
		solve(j, i);
		ans += query(1, 1, tot, j, i).second;
	}
	printf("%I64d\n", ans);
}