Codeforces Round钻368(Div.2)E.Garlands二次元樹状配列

テーマリンク
Like all children,Alesha loves New Year cellebration.During the cell ebration he and his whole family dress up the fir-tree.Like alldren,Alesha likes to Playwith garlands — chains consisting of a lightbulbs.
Alesha uses a grid field sized n𔎅×m for playing.The rows of the field from 1 to n from the top to the bottom and columns are numbend from 1 to m from the the the right.
Aleshas k garlands which he placces the field.He does so in the way such that each lightbulb of each garland lies in the center of some cell in the field,and each cell contains most lightbulb.coarse.
Each garland is turned off or turned on at any moment.If some garland is turned on then each of its lightbulbs is turned on,the same appies for garland turned off.Each lightbulb the whole and serland and ingha,breturning,Alach.described by an integer value.Turned off lightbulbs don't bring Alesha any please.
Alesha can turn garland and off and wants to know the sum of pleasure value which the lightbulbs,plced in the centrs of the cell rectanglar part of the field
Alesha is still very little and can't add big numbers.He extremilly asks you to help him.Input
The first line of the input contains three integers n、m and k(1̵≦𔎅n、𔎅m、̵k≦𔎅2000) — the number of field rows,the number of field columns and the number of garlands placed the field rectively.
Next LINE contains garlands set description in the follwing format:
The first line of a single garland description contains a single integer len(1̵≦𔎅len≦𔎅2000) — the number of lightbulbs in the garland.
Each of the next len lineas contains three integers i、j and w(1̵≦𔎅i𔎅n、1𔎅j≦𔎅m、1𔎅≦𔎅m、1󈙕≦󈙕̵󈙕̵̵ — the coordinans of the cell containing a lightbullb and please Alesha gets from it if it is turned on.The lightbulbs are given the order the ary forming a chain the garland.It is Grantated the burecent
The next line contains single integer q(1̵≦̵qq̵≤𔎅106) — the number of event s in Alesha's game.The next q lineas describes events in chronologcal order.The i-th of the m describes the i-th event in the one of the follo wing formas:

SWITCH i — Alesha turns off i-th garland if it is turned on,or turns it on if it is turned off.It is garanted that 1̵≦𔎅i≦𔎅k.

ASK x 1 y 1 x 2 y 2 — Alesha wants to know the sum of pleasure values the lightbulbs,placd in a rectanglar part of the field.Top-left cell of a part has coordination(x 1,̵y1)and right-bottom cel hars 2.It is garanted that 1̵≦x 1𔎅≦x 2𔎅X2𔎅≦𔎅nand 1𔎅̵y1𔎅y2𔎅y2𔎅≦𔎅m.The is theris
All the numbers in the input are integers.
Please note that the input is quite large,so be careful while using some input ways.In particular,it's not recommanded to use cin codes on C+and class Scanner in codes Java.
Output
For each ASK operation the sum Alesha wants to know in a separate line.Print the answers in chronologcal order.
Example
Input

4 4 3
5
1 1 2
1 2 3
2 2 1
2 1 4
3 1 7
4
1 3 1
2 3 3
2 4 3
1 4 1
7
4 1 1
4 2 9
3 2 8
3 3 3
4 3 4
4 4 1
3 4 1
2
ASK 2 2 3 3
ASK 1 1 4 4

Output

15
52

Input

4 4 1
8
4 1 1
3 1 2
2 1 1
1 1 7
1 2 5
2 2 4
2 3 1
1 3 1
3
ASK 1 1 3 2
SWITCH 1
ASK 1 1 3 2

Output

19
0

题意：n*m的矩形空地，有k条串联的灯，每个灯只能占一格，相邻的灯在相邻的空地上。有q个操作，每次可以开关一个链上的灯，或查询一个矩形的权值和。

思路：当要频繁的对数组元素进行修改，同时又要频繁的查询数组内任一区间元素之和的时候，可以考虑使用树状数组。 这儿用二维树状数组。修改操作可以先记着，等到查询的时候再更新树状数组。

#include
using namespace std;
const int maxn = 2e3+5;
int n, m;
long long C[maxn][maxn];
int lowbit(int x)
{
    return x&(-x);
}
int Add(int x, int y, int d){
    for(int i=x; i<=n; i +=lowbit(i))
        for(int j=y; j<=m; j+=lowbit(j))
            C[i][j] += d;
}

long long Sum(int x, int y){
    long long res = 0;
    for(int i=x; i>0; i-=lowbit(i))
        for(int j=y; j>0; j-=lowbit(j))
            res += C[i][j];
    return res;
}
vector vex[maxn],vey[maxn],vew[maxn];
int main()
{
    int k, num, x, y, z, q;
    char op[10];
    int flag[maxn], light[maxn];
    memset(flag,0,sizeof(flag));
    memset(light,0,sizeof(light));
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=k;i++){
        scanf("%d",&num);
        light[i]=1;
        for(int j=1;j<=num;j++){
            scanf("%d%d%d",&x,&y,&z);
            vex[i].push_back(x);
            vey[i].push_back(y);
            vew[i].push_back(z);
        }
    }
    scanf("%d",&q);
    for(int i=1;i<=q;i++){
        scanf("%s",op);
        if(op[0]=='S'){
            int a;
            scanf("%d",&a);
            flag[a] ^= 1;
        }
        else{
            int a, b,c,d;
            scanf("%d%d%d%d",&a,&b,&c,&d);
            for(int j=1;j<=k;j++){
                if(light[j]==flag[j]) continue;
                if(flag[j]==0){
                    for(int ii=0;ii