bzoj3583: 杰杰的女性朋友

原来写的等比矩阵求和是$log^2{}$的，$log{}$的就把最高次按照从低位到高位分成$log{}$个递增的$2$的整次幂的长度

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>

using namespace std;

template<typename Q> Q &read(Q &x) {
    static char c, f;
    for(f = 0; c = getchar(), !isdigit(c); ) if(c == '-') f = 1;
    for(x = 0; isdigit(c); c = getchar()) x = x * 10 + c - '0';
    if(f) x = -x; return x;
}
template<typename Q> Q read() {
    static Q x; read(x); return x;
}

const int N = 20, mod = 1000000007;

int m; 
struct Matrix {
    int da[N][N];
}I, base;

void addit(int &a, int b) {
    if((a += b) >= mod) a -= mod;
}

typedef long long LL;

Matrix operator * (const Matrix &a, const Matrix &b) {
    Matrix c;
    for(int i = 0; i < m; i++) {
        for(int j = 0; j < m; j++) {
            c.da[i][j] = 0;
            for(int k = 0; k < m; k++) {
                addit(c.da[i][j], (LL) a.da[i][k] * b.da[k][j] % mod);
            }
        }
    }
    return c;
}

Matrix operator + (const Matrix &a, const Matrix &b) {
    Matrix c;
    for(int i = 0; i < m; i++) {
        for(int j = 0; j < m; j++) {
            c.da[i][j] = 0;
            addit(c.da[i][j], a.da[i][j]);
            addit(c.da[i][j], b.da[i][j]);
        }
    }
    return c;
}

Matrix solve(Matrix a, int n) {
    Matrix c = I, v = I, b = a;
    for(; ; ) {
        if(n & 1) c = c + b * v, v = v * a;
        if(!(n >>= 1)) return c;
        b = b * (a + I);
        a = a * a;
    }
}

int in[N][2010], out[2010][N];

int main() {
#ifdef DEBUG
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    
    int n;
    scanf("%d%d", &n, &m);
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            scanf("%d", out[i] + j);
        }
        for(int j = 0; j < m; j++) {
            scanf("%d", in[j] + i);
        }
    }
    
    int q; scanf("%d", &q);
    for(int i = 0; i < m; i++) {
        for(int j = 0; j < m; j++) {
            for(int k = 0; k < n; k++) {
                addit(base.da[i][j], (LL) in[i][k] * out[k][j] % mod);
            }
        }
    }
    int ans[N];
    for(int i = 0; i < m; i++) I.da[i][i] = 1;
    while(q--) {
        int u, v, d; scanf("%d%d%d", &u, &v, &d);
        if(d == 0) printf("%d
", u == v);
        else {
            memset(ans, 0, sizeof ans);
            --u, --v;
            Matrix c = solve(base, d - 1);
            for(int j = 0; j < m; j++) {
                for(int k = 0; k < m; k++) {
                    addit(ans[j], (LL) out[u][k] * c.da[k][j] % mod);
                }
            }
            int res = 0;
            for(int i = 0; i < m; i++) {
                addit(res, (LL) ans[i] * in[i][v] % mod);
            }
            printf("%d
", res + (u == v));
        }
    }
    
    return 0;
}