Codeforces 707E Garlands（二维树状数组）

给你n∗m的矩阵，然后k个花环，每个有leni个灯，然后一开始都是开的

10W次操作，可以把这个花环的所有灯都变一个状态，然后求子矩阵内有多少亮的

求和的操作就2000次

想了个20003的暴力肯定不行

题解也没怎么懂，看lxc巨菊二维树状数组过了，但是算了下复杂度有2000∗2000∗50∗log2000啊

只要记录一下这次求和和上次操作的时候状态一不一样，不一样就修改

玄学复杂度过题。。。就当复习二维树状数组了

代码：

#include <map>
#include <set>
#include <ctime>
#include <stack>
#include <queue>
#include <cmath>
#include <bitset>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#pragma comment(linker,"/STACK:102400000,102400000")

using namespace std;
#define   MAX           2005
#define   MAXN          1000005
#define   maxnode       205
#define   sigma_size    26
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   lrt           rt<<1
#define   rrt           rt<<1|1
#define   middle        int m=(r+l)>>1
#define   LL            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   pii           pair<int,int>
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   limit         10000

//const int    prime = 999983;
const int    INF   = 0x3f3f3f3f;
const LL     INFF  = 0x3f3f;
const double pi    = acos(-1.0);
const double inf   = 1e18;
const double eps   = 1e-4;
const LL    mod    = 1e9+7;
const ull    mx    = 133333331;

/*****************************************************/
inline void RI(int &x) {
char c;
while((c=getchar())<'0' || c>'9');
x=c-'0';
while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
/*****************************************************/
LL c[MAX][MAX];
int n,m;

void add(int x, int y, int w) {
for (int i = x; i <= n; i += lowbit(i)) {
for (int j = y; j <= m; j += lowbit(j)) {
c[i][j] += w;
}
}
}
LL sum(int x, int y) {
LL ret = 0;
for (int i = x; i > 0; i -= lowbit(i)) {
for (int j = y; j > 0; j -= lowbit(j)) {
ret += c[i][j];
}
}
return ret;
}
vector<pair<pii,int> > v[MAX];
int vis[MAX];
int last[MAX];
int main(){
int k;
while(cin>>n>>m>>k){
mem(c,0);
for(int i=1;i<=k;i++){
v[i].clear();
int len;
scanf("%d",&len);
while(len--){
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
v[i].push_back(mk(mk(a,b),c));
add(a,b,c);
}
}
int q;
cin>>q;
mem(last,0);mem(vis,0);
while(q--){
char op[10];
scanf("%s",op);
if(op[0]=='S'){
int a;
scanf("%d",&a);
vis[a]^=1;
}
else{
int x1,y1,x2,y2;
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
for(int i=1;i<=k;i++){
if(last[i]!=vis[i]){
for(int j=0;j<v[i].size();j++){
if(vis[i]) add(v[i][j].first.first,v[i][j].first.second,-v[i][j].second);
else add(v[i][j].first.first,v[i][j].first.second,v[i][j].second);
}
last[i]=vis[i];
}
}
LL ans=sum(x2,y2)-sum(x2,y1-1)-sum(x1-1,y2)+sum(x1-1,y1-1);
cout<<ans<<endl;
}
}
}
return 0;
}