Title:

There is a chessboard with an initial color of all white n*n. each time you select a rectangular area to reverse the color, ask how many black squares there will be at the end. N, k<=10000

Analysis:

It's the same as scanning line. Record each transverse edge and project it to the x-axis. It's just that there's no distinction between the top and the bottom. After sorting by height, for each edge, project to the x-axis, and perform xor 1 operation on the projection interval once. Therefore, the segment tree maintains interval exclusive or operation, and then queries interval sum every time.

Code:

#include<bits/stdc++.h>
#define lson rt<<1,l,(l+r)/2
#define rson rt<<1|1,(l+r)/2+1,r
using namespace std;
const int MAXN = 1e4+5;
int n,k,cnt;
struct Edge {
	int l,r,h;
	bool operator < (const Edge &e) const {
		return h < e.h;
	}
}e[MAXN<<1];
struct Seg_Tree{
	int sum[MAXN << 2], tag[MAXN<<2];
	void build() {
		memset(sum,0,sizeof sum);
		memset(tag,0,sizeof tag);
	}
	void pushup(int rt) {
		sum[rt] = sum[rt<<1] + sum[rt<<1|1];
	}
	void pushdown(int rt,int l,int r) {
		if(tag[rt]){
			tag[rt<<1] ^= tag[rt];
			tag[rt<<1|1] ^= tag[rt];
			sum[rt<<1] = (l+r)/2 - l + 1 - sum[rt<<1];
			sum[rt<<1|1] = r - (l+r)/2 - sum[rt<<1|1];
			tag[rt] = 0;
		}
	}
	void update(int L,int R,int rt,int l,int r) {
		if(L<=l && R>=r){
			tag[rt] ^= 1;
			sum[rt] = (r-l+1) - sum[rt];
			return;
		}
		pushdown(rt,l,r);
		if(L<=(l+r)/2)	update(L,R,lson);
		if(R>(l+r)/2) update(L,R,rson);
		pushup(rt);
	}
}t;
int main() {
	ios::sync_with_stdio(false);
	int T;
	cin >> T;
	while(T--) {
		cin >> n >> k;
		cnt = 0;
		for(int i=0;i<k;i++){
			int xlow,xhigh,ylow,yhigh;
			cin >> xlow >> xhigh >> ylow >> yhigh;
			e[++cnt] = {xlow,xhigh,ylow};
			e[++cnt] = {xlow,xhigh,yhigh+1};
		}
		sort(e+1,e+cnt+1);
		int ans = 0;
		t.build();
		for(int i=1;i<cnt;i++) {
			t.update(e[i].l,e[i].r,1,1,1e4+2);
			ans += t.sum[1]*(e[i+1].h-e[i].h);
		}
		cout << ans << endl;
	}
	return 0;
}