题解 - P4422

Code first, then talk

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

struct node{
	node *ls, *rs;
	int lb, rb, mi;
	node () {
		mi = 0x3f3f3f3f;
		ls = rs = NULL;
	}
};
void build(node *cur, int lb, int rb){
	cur -> lb = lb;
	cur -> rb = rb;
	cur -> mi = 0x3f3f3f3f;
	if (lb != rb){
		cur -> ls = new node;
		cur -> rs = new node;
		int mid = (lb + rb) >> 1;
		build(cur -> ls, lb, mid);
		build(cur -> rs, mid+1, rb);
	}
}
void sett(node *cur, int x, int y){
	if (cur -> lb == x && cur -> rb == x){
		cur -> mi = y;
	}
	else {
		int mid = (cur -> lb + cur -> rb) >> 1;
		if (x <= mid) sett(cur -> ls, x, y);
		else sett(cur -> rs, x, y);
		cur -> mi = min(cur -> ls -> mi, cur -> rs -> mi);
	}
}
int getm(node *cur, int lb, int rb){
	if (cur -> lb == lb && cur -> rb == rb) return cur -> mi;
	int ans = 0x3f3f3f3f;
	int mid = (cur -> lb + cur -> rb) >> 1;
	if (lb <= mid) ans = min(ans, getm(cur -> ls, lb, min(mid, rb)));
	if (rb > mid) ans = min(ans, getm(cur -> rs, max(mid+1, lb), rb));
	return ans;
}
node *root;
int solve(node *cur, int lb, int rb, int x){
	if (cur -> lb == cur -> rb) {
		if (cur -> mi <= x) return cur -> lb;
		else return -1;
	}
	int mid = (cur -> lb + cur -> rb) >> 1;
	if (rb <= mid) return solve(cur -> ls, lb, rb, x);
	else if (lb > mid) return solve(cur -> rs, lb, rb, x);
	else {
		if (getm(cur -> ls, lb, mid) <= x) return solve(cur -> ls, lb, mid, x);
		else return solve(cur -> rs, mid+1, rb, x);
	}
}
int n, q;
int main(){
	scanf("%d%d", &n, &q);
	root = new node;
	build(root, 1, n);
	int x, y;
	int lc, rc;
	lc = rc = 0;
	char c;
	for (int i = 1;i <= q;i++){
		c = getchar();
		while (!isalpha(c)) c = getchar();
		scanf("%d%d", &x, &y);
		if (c == 'M'){
			sett(root, y, x);
		}
		else{
			int anss = solve(root, y, n, x);
			if (anss == 0x3f3f3f3f) printf("-1\n");
			else printf("%d\n", anss);
		}
	}
	return 0;
}

题目思路

这道题一看容易觉得是 mod hist se_tr ，但其实不能这样做。

理清 BF 思路，然后发现其实可以用 se_tr 维护 当前年龄的下车时间。

设年龄为 \(i\) 的人下车时间为 \(i\)，对于输入 \(x, y\) 要求出 \(min_{y\le i\le n, d_i\ge x}(d_i)\) 。

于是线段树上二分就对了。