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，不一定连通性只是在这个区间内的比如虽然区间能从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;const int N = 100010;LL;int n;char s[10];struct Node {int l, r, h0, h1, s0, s1, x0, x1;}a[N << 2];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;}
『 不可能 』只存在於蠢人的字典里
