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  • Codeforces 348 D

     

    #include<bits/stdc++.h>
    using namespace std;
    #define y1 y11
    #define fi first
    #define se second
    #define pi acos(-1.0)
    #define LL long long
    //#define mp make_pair
    #define pb push_back
    #define ls rt<<1, l, m
    #define rs rt<<1|1, m+1, r
    #define ULL unsigned LL
    #define pll pair<LL, LL>
    #define pli pair<LL, int>
    #define pii pair<int, int>
    #define piii pair<pii, int>
    #define pdd pair<double, double>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    //head
     
    const int N = 3e3 + 5;
    const int MOD = 1e9 + 7;
    int dp[N][N], n, m;
    char s[N][N];
    int solve(int a, int b, int c, int d) {
        for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j) dp[i][j] = 0;
        for (int i = a; i <= c; ++i) {
            for (int j = b; j <= d; ++j) {
                if(i == a && j == b) {
                    if(s[i][j] == '.') dp[i][j] = 1;
                }
                else {
                    if(s[i][j] == '.') dp[i][j] = (dp[i-1][j]+dp[i][j-1])%MOD;
                }
            } 
        } 
        return dp[c][d];
    }
    int main() {
        scanf("%d %d", &n, &m);
        for (int i = 1; i <= n; ++i) scanf("%s", s[i]+1);
        printf("%lld
    ", (solve(1, 2, n-1, m)*1LL*solve(2, 1, n, m-1) - solve(1, 2, n, m-1)*1LL*solve(2, 1, n-1, m)%MOD+MOD)%MOD);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Aragaki/p/11773411.html
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