题解 | Maze-2019牛客暑期多校训练营第二场E题

题目来源于牛客竞赛：https://ac.nowcoder.com/acm/contest/discuss
题目描述：
Given a maze with N rows and M columns, where b_ij represents the cell on the i-row, j-th column. If b_ij=“1” , it’s a wall and can’t not be passed. If you are on the cell b_i,j, ou can go to b_i,(j+1) as long as it’s not a wall.

Sometime, a cell may be changed into wall, or vise versa. You need to find out the number of way to pass through the maze starting at some given cell and finishing at some given cell.

If the starting cell or finishing cell is a wall, there’s clearly no way to pass through the maze.

Note that you can’t go back to the cell you just from.

输入描述：
The first line of input contains three space-separated integers N, M, Q.
Following N lines each contains M characters b_ij representing the maze.
Following Q lines each contains three space-separated integers

If q_i=1, the state of cell b_ai,bi is changed.
If q_i=2, you need to find out the number of way to start at cell b_1,ai and finish at cell b_N,bi.

1≤N,Q≤50000
1≤M≤10
b_i,j∈ “01”
1≤q_i≤2
If q_i=1, 1≤a_i≤N and 1≤b_i≤M.
If q_i=2, 1≤a_i,b_i≤M.

输出描述：
For each q_i=2, Output one line containing an integer representing the answer module 10⁹+7(1000000007).

示例1：
输入
2 2 3
00
00
2 1 2
1 1 2
2 1 2

输出
2
1

题解：
• Data Structure(Segment tree), Dynamic Programming
We can construct a DP as:
dp[i][j] =sum(dp[i-1][k] for (k < j and b_ik=b_i{k+1}=…=b_ij=0))+ sum(dp[i-1][k] for (k > j and b_ik=b_i{k-1}=…=b_ij=0))
It can be represented as matrix multiplication from dp[i][ * ] to dp[i+1][ * ].

• Data Structure(Segment tree), Dynamic Programming
dp[i][j] = sum(dp[i-1][k] for (k < j and b_ik=b_i{k+1}=…=b_ij=0))+ sum(dp[i-1][k] for (k > j and b_ik=b_i{k-1}=…=b_ij=0))
Construct a segment tree, each node consists of the weight of each state.
The final answer will be the product of them.
Then, modify time will be O(m^3 \lg N), query time will be O(1).

代码：

// Author: Yen-Jen Wang
#pragma GCC optimize("O3")
#include 

using namespace std;

typedef long long ll;

const int MAX_N = 50000 + 7;
const int MAX_M = 10 + 1;

const ll MOD = 1000000000 + 7;

struct Node {
    Node *lc, *rc;
    int l, r;
    ll dp[MAX_M][MAX_M];

    Node(int _l = 0, int _r = 0) : lc(0), rc(0), l(_l), r(_r) {
        memset(dp, 0, sizeof(dp));
    }
};

int N, M;
char mp[MAX_N][MAX_M];

void pull(Node *o) {
    //static ll dp1[MAX_M];
    //static ll dp2[MAX_M];

    Node *a = o->lc;
    Node *b = o->rc;

    for (int i = 1; i <= M; ++i) {
      for (int j = 1; j <= M; ++j) {
        o->dp[i][j] = 0;
      }
    }

    for (int i = 1; i <= M; ++i) {
      for (int j = 1; j <= M; ++j) {
        if (a->dp[i][j] == 0) {
          continue;
        }
        for (int k = 1; k <= M; ++k) {
          ll way = (a->dp[i][j] * b->dp[j][k]) % MOD;
          o->dp[i][k] = (o->dp[i][k] + way) % MOD;
        }
      }
    }

    /*for (int i = 1; i <= M; i++) {
        for (int j = 1; j <= M; j++) {
            if (!a->dp[i][j])
                continue;
            if (mp[b->l][j]) 
                dp1[j] = a->dp[i][j];
            else
                dp1[j] = 0;
        }
        
        for (int j = 1; j <= M; j++) 
            dp2[j] = 0;
        
        ll s = 0;
        for (int j = 1; j <= M; j++) {
            if (mp[b->l][j] == 0)
                s = 0;
            else {
                s += dp1[j];
                if (s >= MOD)
                    s -= MOD;
            }
            dp2[j] += s;
            if (dp2[j] >= MOD)
                dp2[j] -= MOD;
        }

        s = 0;
        for (int j = M; j >= 0; j--) {
            if (mp[b->l][j] == 0)
                s = 0;
            dp2[j] += s;
            if (dp2[j] >= MOD)
                dp2[j] -= MOD;
            if (mp[b->l][j] != 0) {
                s += dp1[j];
                if (s >= MOD)
                    s -= MOD;
            }
        }

        for (int j = 1; j <= M; j++) {
            o->dp[i][j] = 0;
            for (int k = 1; k <= M; k++) {
                o->dp[i][j] += dp2[k] * b->dp[k][j] % MOD;
                if (o->dp[i][j] >= MOD)
                    o->dp[i][j] -= MOD;
            }
        }
    }*/
}

Node* build(int l, int r) {
    Node *o = new Node(l, r);
    if (l == r) {
        for (int i = 1; i <= M; i++) {
          for (int j = 1; j <= M; j++) {
            o->dp[i][j] = 0;
          }
          for (int j = i; j <= M; ++j) {
            if (mp[l][j] == 0) {
              break;
            }
            o->dp[i][j] = 1;
          }
          for (int j = i; j >= 1; --j) {
            if (mp[l][j] == 0) {
              break;
            }
            o->dp[i][j] = 1;
          }
        }
    }
    else {
        int m = (l + r) >> 1;
        o->lc = build(l, m);
        o->rc = build(m + 1, r);
        pull(o);
    }
    return o;
}

void maintain(Node *o, int p) {
    int l = o->l, r = o->r;
    if (l == r) {
        for (int i = 1; i <= M; i++) {
          for (int j = 1; j <= M; j++) {
            o->dp[i][j] = 0;
          }
          for (int j = i; j <= M; ++j) {
            if (mp[l][j] == 0) {
              break;
            }
            o->dp[i][j] = 1;
          }
          for (int j = i; j >= 1; --j) {
            if (mp[l][j] == 0) {
              break;
            }
            o->dp[i][j] = 1;
          }
        }
    }
    else {
        int m = (l + r) >> 1;
        if (p <= m)
            maintain(o->lc, p);
        else
            maintain(o->rc, p);
        pull(o);
    }
}

int main() {
    int Q;
    scanf("%d%d%d", &N, &M, &Q); 
    for (int i = 1; i <= N; i++) {
        scanf("%s", mp[i] + 1);
        for (int j = 1; j <= M; j++){
            mp[i][j] -= '0';
            mp[i][j] = 1 - mp[i][j];
        }
    }
    Node *tr = build(1, N);
    while (Q--) {
        int q, a, b;
        scanf("%d%d%d", &q, &a, &b);
        if (q == 1) {
            mp[a][b] ^= 1;
            maintain(tr, a);
        }
        else 
            printf("%lld\n", tr->dp[a][b]);
    }
    return 0;
}

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