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  • BZOJ3165: [Heoi2013]Segment(李超线段树)

    题意

    题目链接

    Sol

    李超线段树板子题。具体原理就不讲了。

    一开始自己yy着写差点写自闭都快把叉积搬出来了。。。

    后来看了下litble的写法才发现原来可以写的这么清晰简洁Orz

    #include<bits/stdc++.h>
    #define pdd pair<double, double> 
    #define MP make_pair
    #define fi first
    #define se second 
    using namespace std;
    const int MAXN = 1e6 + 10, Lim = 1e9;
    template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
    template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
    inline int read() {
    	char c = getchar(); int x = 0, f = 1;
    	while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    	return x * f;
    }
    int N = 39989, M;
    int ls[MAXN], rs[MAXN], root, cnt, tot;
    pdd mx[MAXN];
    struct Line {
    	double k, b;
    	int id;
    }s[MAXN];
    pdd get(int x0, int y0, int x1, int y1) {
    	double k = (double) (y1 - y0) / (x1 - x0),
    		   b = (double) y0 - k * x0;
    	return {k, b};
    }
    double calc(Line line, int x) {
    	return line.k * x + line.b;
    }
    double GetPoint(Line a, Line b) {
    	return (b.b - a.b) / (a.k - b.k);
    }
    pdd ret;
    void Query(int k, int l, int r, int p) {//fi: val  se: id
    	if(chmax(ret.fi, calc(s[k], p))) ret.se = s[k].id;
    	if(l == r) return ;
    	int mid = l + r >> 1;
    	if(p <= mid) Query(ls[k], l, mid, p);
    	else Query(rs[k], mid + 1, r, p);
    }
    void Modify(int &k, int l, int r, int ql, int qr, Line seg) {
    	if(!k) k = ++tot;
    	int mid = l + r >> 1;
    	if(ql <= l && r <= qr) {
    		if(!s[k].id) {s[k] = seg; return ;}
    		int p = GetPoint(s[k], seg);
    		int pl = calc(s[k], l), pr = calc(s[k], r), nl = calc(seg, l), nr = calc(seg, r);
    		if(pl > nl && pr > nr) return ;
    		if(pl < nl && pr < nr) {s[k] = seg; return ;}
    		if(pl < nl) {
    			if(p > mid) Modify(rs[k], mid + 1, r, mid + 1, r, s[k]), s[k] = seg;
    			else Modify(ls[k], l, mid, l, mid, seg);
    		} else {
    			if(p > mid) Modify(rs[k], mid + 1, r, mid + 1, r, seg);
    			else Modify(ls[k], l, mid, l, mid, s[k]), s[k] = seg;
    		}
    		return ;
    	}
    	if(l == r) return ;
    	if(ql <= mid) Modify(ls[k], l, mid, ql, qr, seg);
    	if(qr  > mid) Modify(rs[k], mid + 1, r, ql, qr, seg);
    }
    signed main() {
    	M = read();
    	for(int i = 1, lastans = 0; i <= M; i++) {
    		int opt = read();
    		if(!opt) {
    			int k = read(), x = (k + lastans - 1) % 39989 + 1;
    			ret.fi = 0; ret.se = 0;
    			Query(root, 1, N, x);
    			printf("%d
    ", lastans = (mx[x].fi > ret.fi ? mx[x].se : ret.se));
    		} else {
    			int x0 = (read() + lastans - 1) % 39989 + 1, y0 = (read() + lastans - 1) % Lim + 1,
    				x1 = (read() + lastans - 1) % 39989 + 1, y1 = (read() + lastans - 1) % Lim + 1;
    			if(x0 > x1) swap(x0, x1), swap(y0, y1);
    			if(x0 == x1 && chmax(mx[x0].fi, max(y0, y1))) mx[x0].se = i;
    			pdd li = get(x0, y0, x1, y1);
    			Modify(root, 1, N, x0, x1, {li.fi, li.se, ++cnt});
    		}
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/10355576.html
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