POJ 3695 rectangles 1w ask for 20 matrix area and

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The main idea is to give a number of rectangles (n <= 20) and then M queries (M <= 100000)

Each query gives a number of rectangles, asking how large the area of these rectangles is.

When it comes to rectangles, maybe the first reaction is a line tree.

But this problem has a feature, that is, n is very small, M is very large

It's probably not going to work with a line tree.

So change the idea, n is very small, we can consider all possible combinations, and then preprocessing out, so the inquiry is O (1) query

But 1<<20 is clearly more than 100000.

That means we don't have to think about all of this.

You just have to think about the situation in M-queries.

So we first read the situation in the inquiry in, with the binary system to save up.

Then there's DFS, according to the principle of repulsion.

The area of a rectangle-two rectangles intersect the area + three rectangles intersect the area ... That's it

So DFS can have two branches, one is to take this rectangle, the other is not to take

#include <stdio.h> #include <iostream> #include <algorithm> #include <sstream> #include < stdlib.h> #include <string.h> #include <limits.h> #include <vector> #include <string># Include <time.h> #include <math.h> #include <queue> #include <stack> #include <set># Include <map>const int inf = 1e8;const double eps = 1e-8;const double pi = ACOs ( -1.0); template <class t> Inli      ne BOOL Rd (T &ret) {char c; int sgn;      if (C=getchar (), c==eof) return 0; while (c!= '-' && (c< ' 0 ' | |      C> ' 9 ')) C=getchar ();      sgn= (c== '-')? -1:1;      ret= (c== '-')? 0: (c ' 0 ');      while (C=getchar (), c>= ' 0 ' &&c<= ' 9 ') ret=ret*10+ (c ' 0 ');      RET*=SGN;  return 1;      } template <class t> inline void pt (T x) {if (x <0) {Putchar ('-'); x =-X;}      if (x>9) pt (X/10);  Putchar (x%10+ ' 0 ');    }using namespace Std;typedef Long long ll;typedef pair<int,int> PII; int n, m; struct NODE{INT X1, x2, y1, Y2;int area () {return abs (X1-X2) *abs (y1-y2);}} A[30];int ask[100005];int st[1<<21];void dfs (int xa, int ya, int xb, int yb, int deep, int flag, int sta) {if (Xa > = XB | | Ya >= yb) return;if (deep = = N) {if (STA) {(int i = 1; I <= m; i++) if ((ask[i]|sta) = = Ask[i]) st[ask[i]] + = flag* (xb-xa ) * (Yb-ya);} return;} DFS (xa, ya, xb, YB, deep+1, Flag, STA);d FS (max (XA, a[deep+1].x1), Max (ya, a[deep+1].y1), min (XB, a[deep+1].x2), Min (YB, a[ Deep+1].y2), deep+1,-flag, sta| (1<<deep));}    int main () {int Cas = 1;  while (Cin>>n>>m, n+m) {printf (' Case%d:\n ', cas++), for (int i = 1; I <= n; i++) {rd (a[i].x1); Rd (A[I].Y1); RD (A[I].X2); RD (A[I].Y2);} for (int i = 1, siz, num; i <= m; i++) {Ask[i] = 0;rd (siz); while (siz-->0) {rd (num); ask[i]|=1<< (num-1);}} memset (St, 0, sizeof St);d FS (0,0,inf,inf, 0,-1,0), for (int i = 1; I <= m; i++) printf ("Query%d:%d\n", I, St[ask[i]]);p u    TS ("");} return 0;}

POJ 3695 rectangles 1w ask for 20 matrix area and