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  • Luogu P4097 [HEOI2013]Segment 李超线段树

    题目链接 (Click) (Here)

    李超线段树的模板。但是因为我实在太(Naive)了,想象不到实现方法。

    看代码就能懂的东西,放在这里用于复习。

    #include <bits/stdc++.h>
    using namespace std;
    
    const int N = 100010;
    #define ls (p << 1)
    #define rs (p << 1 | 1)
    #define mid ((l + r) >> 1)
    				 
    struct Node {
        int l, r, id;
        double yl, yr;
        Node (int x1 = 0, int y1 = 0, int x2 = 0, int y2 = 0, int i = 0) {
            l = x1, r = x2, yl = y1, yr = y2, id = i;
            if (l == r) {
    			yl = yr = max (yl, yr);
    		}
        }
        double get (int x) {return l == r ? yl : yl + (k () * (x - l));}
        double k () {return (yr - yl) / (r - l);}
        void lm (int x) {yl = get (x); l = x;}
        void rm (int x) {yr = get (x); r = x;}
    };
    
    bool hei (Node a, Node b, int x) {
        return a.get (x) == b.get (x) ? a.id < b.id : a.get (x) > b.get (x);
    }
    
    struct St {
        Node tree[N << 2];
    
    	void build (int l, int r, int p) {
            tree[p].l = l;
    		tree[p].r = r;
            if (l == r) return;
            build (l, mid, ls);
            build (mid + 1, r, rs); 
        }
    	
        Node query (int t, int l, int r, int p) {
            if (l == r) return tree[p];
    		Node res;
            if (t <= mid) {
    			res = query (t, l, mid, ls);
            } else {
    			res = query (t, mid + 1, r, rs);
    		}
    		return hei (res, tree[p], t) ? res : tree[p];
        }
    	
        void update (int l, int r, Node k, int p) {
            if (tree[p].l > k.l) k.lm (tree[p].l);
            if (tree[p].r < k.r) k.rm (tree[p].r); //削足适履
            if (hei (k, tree[p], mid)) swap (tree[p], k); //让tree[p]在mid上具有优势
            if (min (tree[p].yl, tree[p].yr) >= max (k.yl, k.yr)) return; //如果完全覆盖
            if (l == r) return; //如果大小为1
            if (tree[p].k () <= k.k ()) {
    			update (mid + 1, r, k, rs); //如果k在后面有露出来的情况
            } else {
    			update (l, mid, k, ls); //如果k在前面有露出来的情况
    		}
    	}
    	
        void insert (int l, int r, Node k, int p) {
            if (k.l > r || k.r < l) return;
            if (tree[p].l > k.l) k.lm (tree[p].l);
            if (tree[p].r < k.r) k.rm (tree[p].r);
            if (l == k.l && r == k.r) {
    			update (l, r, k, p);
    			return;
    		}
    		//把node一路传下去,对应区间就削成对应大小的线段
            if (l == r) return;
    		insert (l, mid, k, ls);
    		insert (mid + 1, r, k, rs);
        }
    }T;
    
    const int My = 1e9;
    const int Mx = 39989;
    
    int m, k, la, Ind, opt;
    
    int main () {
        T.build (1, Mx, 1);
        cin >> m;
        while (m--) {
            cin >> opt;
            if (opt == 0) {
                cin >> k;
                k = (k + la - 1) % Mx + 1;
                la = T.query (k, 1, Mx, 1).id;
                cout << la << endl;
            } else {
                int x0, x1, y0, y1;
                cin >> x0 >> y0 >> x1 >> y1;
                x0 = (x0 + la - 1) % Mx + 1;
    			x1 = (x1 + la - 1) % Mx + 1;
                y0 = (y0 + la - 1) % My + 1;
    			y1 = (y1 + la - 1) % My + 1;
                if (x0 > x1) {
    				swap (x0, x1);
    				swap (y0, y1);
    			}
                Node res = Node (x0, y0, x1, y1, ++Ind);
                T.insert (1, Mx, res, 1);
            }
        }
    }
    
    
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  • 原文地址:https://www.cnblogs.com/maomao9173/p/10444517.html
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