CF1523G Try Booking 题解

link

Solution

我好蠢啊。。。

考虑如果x确定怎么办？我们可以直接递归，每次处理 \([l,r]\) 的最大贡献，然后找到在这个区间里面编号最小的区间，然后两边递归下去。考虑这样的复杂度，因为最多 \(\lfloor \frac{n}{x} \rfloor\) 个区间，所以复杂度是 \(\lfloor\frac{n}{x}\rfloor\log^2 n\) 的。

然后我们直接对于每个 \(x\) 做一遍就好了。3200 的清新题目。呜呜。。

Code

#include <bits/stdc++.h>
using namespace std;

#define Int register int
#define MAXN 100005

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
template <typename T> inline void chkmax (T &a,T b){a = max (a,b);}
template <typename T> inline void chkmin (T &a,T b){a = min (a,b);}

#define LOGN 205
#define ls(x) son[x][0]
#define rs(x) son[x][1]
int n,m,cnt,rt[MAXN],sl[MAXN],sr[MAXN],ind[MAXN],minv[MAXN * LOGN],son[MAXN * LOGN][2];

void modify (int &x,int l,int r,int pos,int miv){
	if (!x) x = ++ cnt,minv[x] = 1e9;
	chkmin (minv[x],miv);
	if (l == r) return ;
	int mid = l + r >> 1;
	if (pos <= mid) modify (ls(x),l,mid,pos,miv);
	else modify (rs(x),mid + 1,r,pos,miv);
}

int query (int x,int l,int r,int ql,int qr){
	if (!x) return 1e9;
	if (l >= ql && r <= qr) return minv[x];
	int mid = l + r >> 1,res = 1e9;
	if (ql <= mid) chkmin (res,query (ls(x),l,mid,ql,qr));
	if (qr > mid) chkmin (res,query (rs(x),mid + 1,r,ql,qr));
	return res;
}

int lowbit (int x){return x & (-x);}

void Modify (int l,int r,int p){
	for (;r <= n;r += lowbit (r)) modify (rt[r],1,n,l,p);
}

int Query (int l,int r){
	int res = 1e9;
	for (;r >= l;r -= lowbit (r)) chkmin (res,query (rt[r],1,n,l,n));
	return res;
}

int solve (int l,int r){
	if (l > r) return 0;
	int res = Query (l,r);
	if (res == 1e9) return 0;
	return sr[res] - sl[res] + 1 + solve (l,sl[res] - 1) + solve (sr[res] + 1,r);
}

int ans[MAXN];

signed main(){
	read (n,m);
	for (Int i = 1;i <= m;++ i) read (sl[i],sr[i]),ind[i] = i;
	sort (ind + 1,ind + m + 1,[](int x,int y){return sr[x] - sl[x] > sr[y] - sl[y];});
	for (Int x = n,i = 1;x >= 1;-- x){
		while (i <= m && sr[ind[i]] - sl[ind[i]] + 1 == x) Modify (sl[ind[i]],sr[ind[i]],ind[i]),i ++;
		ans[x] = solve (1,n);
	}
	for (Int x = 1;x <= n;++ x) write (ans[x]),putchar ('\n');
	return 0;
}