luogu P3801 红色的幻想乡

嘟嘟嘟

首先人人都能想到是线段树，不过二维线段树肯定会MLE+TLE的。

我们换一种想法，不去修改整个区间，而是修改一个点：开横竖两个线段树，分别记录哪些行和列被修改了。因为如果两阵红雾碰撞，则会因为密度过大而沉降消失，所以自然想到亦或。

统计的时候求出这个矩形内有哪些行和列被修改了，接着把这些行和列的长度累加到答案中，但这样的话交点处不仅会重加，而实际上应该是0，所以在减去两倍的被修改的行，列数。

综上：单点亦或修改，区间查询。

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<vector>
 #include<stack>
 #include<queue>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a, x) memset(a, x, sizeof(a))
 #define rg register
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = 1e5 + ;
 inline ll read()
 {
     ll ans = ;
     char ch = getchar(), last = ' ';
     while(!isdigit(ch)) {last = ch; ch = getchar();}
     while(isdigit(ch)) {ans = ans *  + ch - ''; ch = getchar();}
     if(last == '-') ans = -ans;
     return ans;
 }
 inline void write(ll x)
 {
     if(x < ) x = -x, putchar('-');
     if(x >= ) write(x / );
     putchar(x %  + '');
 }

 int n, m, q;
 struct Tree
 {
     int l[maxn << ], r[maxn << ], sum[maxn << ];
     Tree()
     {
         Mem(l, ); Mem(r, );
         Mem(sum, );
     }
     void build(int L, int R, int now)
     {
         l[now] = L; r[now] = R;
         if(L == R) return;
         int mid = (L + R) >> ;
         build(L, mid, now << );
         build(mid + , R, now <<  | );
     }
     void update(int idx, int now)
     {
         if(l[now] == r[now]) {sum[now] ^= ; return;}
         int mid = (l[now] + r[now]) >> ;
         if(idx <= mid) update(idx, now << );
         else update(idx, now <<  | );
         sum[now] = sum[now << ] + sum[now <<  | ];
     }
     int query(int L, int R, int now)
     {
         if(L == l[now] && R == r[now]) return sum[now];
         int mid = (l[now] + r[now]) >> ;
         if(R <= mid) return query(L, R, now << );
         else if(L > mid) return query(L, R, now <<  | );
         else return query(L, mid, now << ) + query(mid + , R, now <<  | );
     }
 }X, Y;

 int main()
 {
     n = read(); m = read(); q = read();
     X.build(, n, ); Y.build(, m, );
     for(int i = ; i <= q; ++i)
     {
         int d = read();
         if(d == )
         {
             int x = read(), y = read();
             X.update(x, ); Y.update(y, );
         }
         else
         {
             int xa = read(), ya = read(), xb = read(), yb = read();
             int a = X.query(xa, xb, ), b = Y.query(ya, yb, );
             write((ll)a * (ll)(yb - ya + ) + (ll)b * (ll)(xb - xa + ) - (ll)a * b * );
             enter;
         }
     }
     return ;
 }