【CSP模拟赛】God knows （李超线段树）

题面

CODE

稍微分析一下，发现把$(i,p_i)$看做二维数点，就是求极长上升子序列的权值最小值。

直接李超线段树

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200005;
const int INF = 0x3f3f3f3f;
int n, v, mxr, p[MAXN];
int mx[MAXN<<2], vl[MAXN<<2], val[MAXN<<2];
int cal(int i, int l, int r, int R) {
	if(l == r) return mx[i] > R ? val[i] : INF;
	int mid = (l + r) >> 1;
	if(mx[i<<1|1] >= R) return min(vl[i], cal(i<<1|1, mid+1, r, R));
	else return cal(i<<1, l, mid, R);
}
void insert(int i, int l, int r, int x, int pos, int V) {
	if(l == r) { mx[i] = pos; val[i] = V; return; }
	int mid = (l + r) >> 1;
	if(x <= mid) insert(i<<1, l, mid, x, pos, V);
	else insert(i<<1|1, mid+1, r, x, pos, V);
	mx[i] = max(mx[i<<1], mx[i<<1|1]);
	vl[i] = cal(i<<1, l, mid, mx[i<<1|1]);
}
void query(int i, int l, int r, int x) {
	if(r <= x) { v = min(v, cal(i, l, r, mxr)); mxr = max(mxr, mx[i]); return; }
	int mid = (l + r) >> 1;
	if(x > mid) query(i<<1|1, mid+1, r, x);
	query(i<<1, l, mid, x);
}
int main () {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) scanf("%d", &p[i]);
	memset(vl, 0x3f, sizeof vl);
	for(int i = 1, c; i <= n; ++i) {
		scanf("%d", &c);
		v = INF, mxr = 0, query(1, 1, n, p[i]);
		insert(1, 1, n, p[i], i, (v == INF ? 0 : v) + c);
	}
	v = INF, mxr = 0, query(1, 1, n, n);
	printf("%d
", v);
}