P3801 red fantasy township

Basic Line Segment tree and rejection Principle

Good food for me!

In fact, there was a slight turn in the question: she separated the two fog sets and settled and placed them in the same place.

However, you only need to draw a graph and you will find that, once there is a cross between the marked rows and columns, this point is useless.

Putting fog on one point is equivalent to putting fog on one line and one column.

The fog in a region is equivalent to the total area occupied by all marked rows and columns and then the area that falls.

It is not hard to find: when it is in the row\ (X \)Fog, put in the column\ (Y \)When it's foggy\ (Xy \).

In fact, it is not difficult to calculate it. You will know it at a glance.

Code:

# Include <cstdio> const int maxn = 100005; struct segtree {int sum [maxn <2]; # define lson (root <1) # define rson (root <1 | 1) void pushup (INT root) {sum [root] = sum [lson] + sum [rson];} void Update (INT root, int L, int R, int p) {If (L = r) sum [root] ^ = 1; else {int mid = (L + r)> 1; if (P <= mid) Update (lson, L, mid, P); else Update (rson, Mid + 1, R, P); pushup (Root );}} int query (INT root, int L, int R, int X, int y) {If (r <X | Y <L) return 0; if (x <= L & R <= y) return sum [root]; int mid = (L + r)> 1; return query (lson, L, mid, x, Y) + query (rson, Mid + 1, R, x, y) ;}# UNDEF lson # UNDEF rson} seg1, seg2; int n, m, Q; int read () {int ans = 0, S = 1; char CH = getchar (); While (CH> '9' | ch <'0 ') {If (CH = '-') S =-1; CH = getchar () ;}while (CH> = '0' & Ch <= '9 ') ans = ans * 10 + CH-'0', CH = getchar (); Return S * ans;} int main () {n = read (), M = read (), q = read (); While (Q --) {int opt = read (); If (OPT = 1) {int x = read (), y = read (); seg1.update (1, 1, n, x); seg2.update (1, 1, n, Y);} else if (OPT = 2) {int x = read (), Y = read (), xx = read (), YY = read (); int RES1 = seg1.query (1, 1, n, x, XX); int RES2 = seg2.query (1, 1, n, Y, YY); printf ("% LLD \ n", 1ll * RES1 * (yy-y + 1) + 1ll * RES2 * (XX-x + 1)-2 * RES1 * RES2);} return 0 ;}

p3801 red fantasy township