题外话

做了这个线段树的题我整个人都不好了，头一次做这种用线段树维护连通性的题，简直烦的要死= =

Description

给你一个2*n的格子，开始全不联通，相邻两点可以连边，有3种操作

1:将(x1,y1),(x2,y2)变为连通

2:将(x1,y1),(x2,y2)变为不连通

3:询问(x1,y1),(x2,y2)是否联通

Soluion

用线段树维护6个信息，正常的4条边以及两条对角线是否联通，修改时用线段树维护

最蛋疼的是查询，比如 [l,r]区间 ，不一定连通性只是在这个区间内的

比如虽然 [l,r] ，虽然不连通，但是可以在 [1,l] 区间能从l上走到下，则l这个位置上下就是联通的，右侧同理，细心处理就可以了

Code

#include 
   
    //good #define ls (rt << 1) #define rs (rt << 1 | 1) using namespace std; const int N = 100010; typedef long long LL; int n; char s[10]; struct Node { int l, r, h0, h1, s0, s1, x0, x1; }a[N << 2]; int g[2][N][4];//U D L R inline int read(int &t) { int f = 1;char c; while (c = getchar(), c < '0' || c > '9') if (c == '-') f = -1; t = c - '0'; while (c = getchar(), c >= '0' && c <= '9') t = t * 10 + c - '0'; t *= f; } Node merge(Node a, Node b) { Node c; int f0 = g[0][a.r][3], f1 = g[1][a.r][3]; c.h0 = (a.h0 && f0 && b.h0) | (a.x0 && f1 && b.x1); c.h1 = (a.h1 && f1 && b.h1) | (a.x1 && f0 && b.x0); c.s0 = a.s0 | (a.h0 && f0 && b.s0 && f1 && a.h1); c.s1 = b.s1 | (b.h0 && f0 && a.s1 && f1 && b.h1); c.x0 = (a.h0 && f0 && b.x0) | (a.x0 && f1 && b.h1); c.x1 = (a.h1 && f1 && b.x1) | (a.x1 && f0 && b.h0); c.l = a.l, c.r = b.r; return c; } void modify(int x1, int y1, int x2, int y2, int x) { if (x1 == x2) g[x1][y1][3] = g[x1][y2][2] = x; else g[0][y1][1] = g[1][y1][0] = x; } void build(int rt, int l, int r) { a[rt].l = l, a[rt].r = r; if (l == r) { a[rt].h0 = a[rt].h1 = 1; return; } int mid = l + r >> 1; build(ls, l, mid), build(rs, mid + 1, r); } void change(int rt, int l, int r, int x1, int y1, int x2, int y2) { int mid = l + r >> 1; if (x1 == x2 && mid == y1) { a[rt] = merge(a[ls], a[rs]); return; } if (l == r && y1 == y2) { a[rt].x0 = a[rt].x1 = 0; a[rt].s0 = a[rt].s1 = g[0][y1][1]; if (a[rt].s0) a[rt].x0 = a[rt].x1 = 1; return ; } if (y1 <= mid) change(ls, l, mid, x1, y1, x2, y2); else change(rs, mid + 1, r, x1, y1, x2, y2); a[rt] = merge(a[ls], a[rs]); } Node find(int rt, int l, int r) { if (l <= a[rt].l && a[rt].r <= r) return a[rt]; int mid = a[rt].l + a[rt].r >> 1; if (r <= mid) return find(ls, l, r); if (l > mid) return find(rs, l, r); else return merge(find(ls, l, r), find(rs, l, r)); } bool ask(int x1, int y1, int x2, int y2) { Node a = find(1, 1, y1), b = find(1, y1, y2), c = find(1, y2, n); b.s0 |= a.s1, b.s1 |= c.s0; if (x1 == x2) { if (x1 == 0) return b.h0 | (b.s0 && b.h1 && b.s1) | (b.s0 && b.x1) | (b.x0 && b.s1); else return b.h1 | (b.s0 && b.h0 && b.s1) | (b.s0 && b.x0) | (b.x1 && b.s1); } else { if (x1 < x2) return b.x0 | (b.h0 && b.s1) | (b.s0 && b.h1); else return b.x1 | (b.h1 && b.s1) | (b.s0 && b.h0); } } int main(){ read(n); build(1, 1, n); while (scanf("%s", s) && s[0] != 'E') { int x1, y1, x2, y2; read(x1), read(y1), read(x2), read(y2); --x1, --x2; if (s[0] == 'C') { if (y1 > y2) swap(y1, y2); modify(x1, y1, x2, y2, 0); change(1, 1, n, x1, y1, x2, y2); } else if (s[0] == 'O') { if (y1 > y2) swap(y1, y2); modify(x1, y1, x2, y2, 1); change(1, 1, n, x1, y1, x2, y2); } else { if (y1 > y2) swap(x1, x2), swap(y1, y2); if (ask(x1, y1, x2, y2)) puts("Y"); else puts("N"); } } return 0; }