题意：

n * m的矩阵，为0表示可以走，1不可以走。规定每走一步只能向下、向左、向右走。现给定两种操作：
一.1 x y表示翻转坐标（x，y）的0、1。
二.2 x y表示从（1，x）走到（n，y）有几种走法

思路：

假设\(dp[i][j]\)表示从下一层能到达（i，j）点的路径数，那么显然到达（i，j）的路径数为\(dp[i + 1][j]\)。
我们能很显然的得到转移方程\(dp[i][j] = \sum_{k = l}^r dp[i - 1][k]\)，其中l~r为（i，j）下方能直接走到（i，j）的连续区间。
我们可以直接用矩阵维护这个转移方程：
\[ \left( \begin{matrix} dp[i][1] & dp[i][2] & dp[i][3] & \cdots & dp[i][m] \end{matrix} \right) * A_i= \left( \begin{matrix} dp[i + 1][1] & dp[i + 1][2] & dp[i + 1][3] & \cdots & dp[i + 1][m] \end{matrix} \right) \]
然后用线段树维护矩阵乘积即可
从（1，x）走到（n，y）只需把x位置置为1，然后乘以\(\prod_{i = 1}^n A_i\)

代码：

#include<cstdio>
#include<set>
#include<cmath>
#include<stack>
#include<vector>
#include<queue>
#include<cstring>
#include<string>
#include<sstream>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn = 50000 + 5;
const int INF = 0x3f3f3f3f;
const ll MOD = 1e9 + 7;
int n, m, q;
int mp[maxn][12];
char s[12];
struct Mat{ll s[12][12];void init_zero(){for(int i = 0; i < 12; i++)for(int j = 0; j < 12; j++)s[i][j] = 0;}
};
Mat pmul(Mat a, Mat b, int len){Mat c;c.init_zero();for(int i = 1; i <= len; i++){for(int j = 1; j <= len; j++){for(int k = 1; k <= len; k++){c.s[i][j] = (c.s[i][j] + a.s[i][k] * b.s[k][j]) % MOD;}}}return c;
}Mat mul[maxn << 2], a[maxn];
void pushup(int rt){mul[rt] = pmul(mul[rt << 1], mul[rt << 1 | 1], m);
}
void build(int l, int r, int rt){if(l == r){for(int i = 1; i <= m; i++)for(int j = 1; j <= m; j++)mul[rt].s[i][j] = a[l].s[i][j];return;}int m = (l + r) >> 1;build(l, m, rt << 1);build(m + 1, r, rt << 1 | 1);pushup(rt);
}
void update(int pos, int l, int r, Mat aa, int rt){if(l == r){for(int i = 1; i <= m; i++)for(int j = 1; j <= m; j++)mul[rt].s[i][j] = aa.s[i][j];return;}int m = (l + r) >> 1;if(pos <= m)update(pos, l, m, aa, rt << 1);elseupdate(pos, m + 1, r, aa, rt << 1 | 1);pushup(rt);
}
int main(){scanf("%d%d%d", &n, &m, &q);for(int i = 1; i <= n; i++){scanf("%s", s + 1);for(int j = 1; j <= m; j++){mp[i][j] = s[j] - '0';}}for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){int base;base = 1;for(int k = j; k >= 1; k--){if(mp[i][k] == 1) base = 0;a[i].s[k][j] = base;}base = 1;for(int k = j; k <= m; k++){if(mp[i][k] == 1) base = 0;a[i].s[k][j] = base;}}}//    for(int k = 1; k <= n; k++){
//        for(int i = 1; i <= m; i++){
//            for(int j = 1; j <= m; j++){
//                printf("%d ", a[k].s[i][j]);
//            }
//            puts("");
//        }
//        puts("*****");
//    }build(1, n, 1);while(q--){int ques, i, j;scanf("%d", &ques);scanf("%d%d", &i, &j);if(ques == 1){mp[i][j] = !mp[i][j];for(int j = 1; j <= m; j++){int base;base = 1;for(int k = j; k >= 1; k--){if(mp[i][k] == 1) base = 0;a[i].s[k][j] = base;}base = 1;for(int k = j; k <= m; k++){if(mp[i][k] == 1) base = 0;a[i].s[k][j] = base;}}update(i, 1, n, a[i], 1);}else{Mat ret;ret.init_zero();ret.s[1][i] = 1;ret = pmul(ret, mul[1], m);printf("%lld\n", ret.s[1][j]);}}return 0;
}
/*
2 6 1
0 0 0 1 0 0
1 0 1 0 1 0
*/