GMOJ5673 爬山法 题解

Solution

显然先想到处理出每个点能看到的最高的顶点。

然后考虑模拟题目的过程，一段一段走时间复杂度显然不够优秀。

考虑我们要求什么，我们需要求出\(u\)到\(v\)的最近的一个点，使得这个点能看到的点比\(v\)能看到的点更高。

然后这个东西可以直接线段树，当然也可以二分+st表

复杂度\(O(n\log n)\)

Code

#include <cstdio>
#include <iostream>
#define LL long long
#define RE register
#define IN inline
using namespace std;
IN int read() {
	int res = 0; char ch = getchar();
	for(; !isdigit(ch); ch = getchar());
	for(; isdigit(ch); ch = getchar()) res = (res << 1) + (res << 3) + (ch ^ 48);
	return res;
}
int n, le[2000010], ri[2000010], nxt[2000010], st[2000010][20], lg[2000010];
LL f[2000010];
struct Point {
	int x, y;
}p[2000010];
inline double slope(int x, int y) {return 1.0 * (p[x].y - p[y].y) / (p[x].x - p[y].x);}
inline bool cmp(int x, int y) {
	if(x == y) return false;
	if(p[x].y > p[y].y) return true;
	if(p[x].y == p[y].y) return p[x].x > p[y].x;
	return false;
}
int stk[2000010];
inline int pmax(int x, int y) {return cmp(x, y) ? x : y;}
void preWork() {
	int top = 0;
	for(int i = 1; i <= n; ++i) {
		while(top > 1 && slope(stk[top], i) >= slope(stk[top], stk[top - 1])) -- top;
		if(top) le[i] = stk[top]; stk[++top] = i;
	}
	top = 0;
	for(int i = n; i; --i) {
		while(top > 1 && slope(stk[top], i) <= slope(stk[top], stk[top - 1])) -- top;
		if(top) ri[i] = stk[top]; stk[++top] = i;
	}
	for(int i = 1; i <= n; ++i) {
		nxt[i] = pmax(le[i], ri[i]);
		if(cmp(i, nxt[i])) nxt[i] = 0;
		st[i][0] = nxt[i];
	}
	for(int i = 1; i <= lg[n]; ++i)
		for(int j = 1; j <= n; ++j)
			st[j][i] = pmax(st[j][i - 1], st[j + (1 << i - 1)][i - 1]);
	return ;
}
inline LL abs(int x) {return x < 0 ? -x : x;}
inline int query(int l, int r) {int len = r - l + 1; return pmax(st[l][lg[len]], st[r - (1 << lg[len]) + 1][lg[len]]);}
LL solve(int k) {
	if(f[k] || !nxt[k]) return f[k];
	int res = nxt[k];
	if(nxt[k] < k) {
		for(int l = nxt[k], r = k - 1, mid; l <= r;) {
			mid = l + r >> 1;
			if(cmp(query(mid, k), nxt[k])) res = mid, l = mid + 1;
			else r = mid - 1;
		}
	}
	else {
		for(int l = k + 1, r = nxt[k], mid; l <= r;) {
			mid = l + r >> 1;
			if(cmp(query(k, mid), nxt[k])) res = mid, r = mid - 1;
			else l = mid + 1;
		}
	}
	f[k] = solve(res) + abs(res - k);
	return f[k];
}
void out(int x) {printf("%d %d\n",p[x].x,p[x].y);}
int main() {
	freopen("mountain.in","r",stdin);
	freopen("mountain.out","w",stdout);
	n = read();
	for(int i = 2; i <= n; ++i) lg[i] = lg[i >> 1] + 1;
	for(int i = 1; i <= n; ++i) p[i].x = read(), p[i].y = read();
	preWork();
	for(int i = 1; i <= n; ++i) solve(i);
	for(int i = 1; i <= n; ++i) printf("%lld\n",f[i]);
	return 0;
}```