More than 2019 cattle off summer school camp (second field) E.MAZE (segment tree + dp)

Meaning of the questions: to give you a matrix of n * m left to right you can only operate with the lower deck there are two q times one is asked to flip a unit (you can not go walking becomes unavailable become available go away) another is to ask from (1, x) come (n, y) how many programs

Ideas: Title 1e5 and n is only 10 m we can consider the tree line segment to maintain a m * m matrix analog matrix multiplication when the time is equivalent to the number of computing solutions (to manually simulate what) modify the operation is equivalent to a single point update then re-structure matrix

#include <bits/stdc++.h>
using namespace std;
const int N = 5e4+7;
const int inf = 0x3f3f3f3f;
typedef long long ll;
const ll mod = 1e9+7;
ll a[N][12];
int n,m,q;
struct matrix{
    int l,r;
    ll ma[12][12];
};
matrix t[N<<2];
void pushup(int p){
    memset(t[p].ma,0,sizeof(t[p].ma));
    for(int i=1;i<=m;i++)
        for(int j=1;j<=m;j++)
            for(int k=1;k<=m;k++){
                t[p].ma[i][j]=(t[p].ma[i][j]+((t[p<<1].ma[i][k]%mod)*(t[p<<1|1].ma[k][j]%mod))%mod)%mod;   
            }
}
void work(int p,int l){
    memset(t[p].ma,0,sizeof(t[p].ma));
    for(int i=1;i<=m;i++){
        int pos=i;
        while(pos>=1&&a[l][pos]==0){
            t[p].ma[i][pos]=1;
            pos--;
        }
        pos=i;
        while(pos<=m&&a[l][pos]==0){
            t[p].ma[i][pos]=1;
            pos++;
        }
    }
}
void build(int p,int l,int r){
    t[p].l=l; t[p].r=r;
    if(l==r){
        work(p,l);
        return ;
    }
    int mid=(l+r)>>1;
    build(p<<1,l,mid);
    build(p<<1|1,mid+1,r);
    pushup(p);
}
void update(int p,int x){
    if(t[p].l==t[p].r){
        work(p,t[p].l);
        return ;
    }
    int mid=(t[p].l+t[p].r)>>1;
    if(x<=mid) update(p<<1,x);
    else update(p<<1|1,x);
    pushup(p);
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cin>>n>>m>>q;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            char b; cin>>b;
            a[i][j]=(b!='0');
        }
    }   
    build(1,1,n);
    for(int i=1;i<=q;i++){
        int z,x,y; cin>>z>>x>>y;
        if(z==1){
            a[x][y]^=1;
            update(1,x);
        }else{
            cout<<t[1].ma[x][y]<<"\n";
        }
    }
    return 0;
}