大意: 給定n*m矩陣, k條鏈, 鏈上每個點有權值, 每次操作可以關閉或打開一條鏈或詢問一個子矩陣內未關閉的權值和.

關鍵詢問操作比較少, 可以枚舉每條鏈, 暴力算出該條鏈對每個詢問的貢獻. 最後再按詢問順序切換鏈的狀態, 只記錄打開的鏈的貢獻即可.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl ‘\n‘
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head




const int N = 2e3+20, M = 1e6+10;
int n, m, k, q, cnt;
int len[N], x[N][N], y[N][N], w[N][N], vis[N];
char op[M][10];
int t[M], t1[N], t2[N], t3[N], t4[N];
ll c[N][N], ans[N][N];
void upd(int x, int y, int w) {
	for (int i=x; i<=n+1; i+=i&-i) for (int j=y; j<=m+1; j+=j&-j) c[i][j]+=w;
}
ll qry(int x, int y) {
	ll r = 0;
	for (int i=x; i; i^=i&-i) for (int j=y; j; j^=j&-j) r+=c[i][j];
	return r;
}

int main() {
	scanf("%d%d%d", &n, &m, &k);
	REP(i,1,k) {
		scanf("%d", len+i);
		REP(j,1,len[i]) { 
			scanf("%d%d%d",&x[i][j],&y[i][j],&w[i][j]);
			++x[i][j], ++y[i][j];
		}
	}
	scanf("%d", &q);
	REP(i,1,q) {
		scanf("%s", op[i]);
		if (op[i][0]==‘S‘) scanf("%d",t+i);
		else { 
			++cnt;
			scanf("%d%d%d%d",t1+cnt,t2+cnt,t3+cnt,t4+cnt);
			++t1[cnt],++t2[cnt],++t3[cnt],++t4[cnt];
		}
	}
	REP(i,1,k) {
		REP(j,1,len[i]) upd(x[i][j],y[i][j],w[i][j]);
		REP(j,1,cnt) {
			int x1=t1[j],y1=t2[j],x2=t3[j],y2=t4[j];
			ans[i][j] = qry(x2,y2)-qry(x2,y1-1)-qry(x1-1,y2)+qry(x1-1,y1-1);
		}
		REP(j,1,len[i]) upd(x[i][j],y[i][j],-w[i][j]);
	}
	int now = 0;
	REP(i,1,q) {
		if (op[i][0]==‘S‘) vis[t[i]] ^= 1;
		else {
			++now;
			ll r = 0;
			REP(i,1,k) if (!vis[i]) r+=ans[i][now];
			printf("%lld\n", r);
		}
	}
}

Garlands CodeForces - 707E (離線樹狀數組)