码迷,mamicode.com
首页 > 其他好文 > 详细

[hdu5411 CRB and Puzzle]DP,矩阵快速幂

时间:2015-08-21 06:59:49      阅读:268      评论:0      收藏:0      [点我收藏+]

标签:

题意:给一个有向图,从任意点开始,最多走m步,求形成的图案总数。

思路:令dp[i][j]表示走j步最后到达i的方法数,则dp[i][j]=∑dp[k][j-1],其中k表示可以直接到达i的点,答案=∑dp[i][j]。关键在于如何减少状态转移的时间,考虑用矩阵加速。

构造矩阵:D = 技术分享,其中a[i][j]表示有向图,用于状态转移,右边的一列1用于累加答案

则答案=[1,1,...1n+1]*DM-1=∑∑DM-1[i][j],1≤i≤n+1,1≤j≤n+1

PS:封装的ModInt放矩阵的最里面进行运算比直接取模慢了3倍多,因此在性能瓶颈地方尽量用最快的写法

 

#pragma comment(linker, "/STACK:10240000")
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}

const double PI = acos(-1.0);
const int INF = 1e9 + 7;
const double EPS = 1e-12;

/* -------------------------------------------------------------------------------- */

const int maxn = 51;

template<int mod>
struct ModInt {
    const static int MD = mod;
    int x;
    ModInt(ll x = 0): x(x % MD) {}
    int get() { return x; }

    ModInt operator + (const ModInt &that) const { int x0 = x + that.x; return ModInt(x0 < MD? x0 : x0 - MD); }
    ModInt operator - (const ModInt &that) const { int x0 = x - that.x; return ModInt(x0 < MD? x0 + MD : x0); }
    ModInt operator * (const ModInt &that) const { return ModInt((long long)x * that.x % MD); }
    ModInt operator / (const ModInt &that) const { return *this * that.inverse(); }

    ModInt operator += (const ModInt &that) { x += that.x; if (x >= MD) x -= MD; }
    ModInt operator -= (const ModInt &that) { x -= that.x; if (x < 0) x += MD; }
    ModInt operator *= (const ModInt &that) { x = (long long)x * that.x % MD; }
    ModInt operator /= (const ModInt &that) { *this = *this / that; }

    ModInt inverse() const {
        int a = x, b = MD, u = 1, v = 0;
        while(b) {
            int t = a / b;
            a -= t * b; std::swap(a, b);
            u -= t * v; std::swap(u, v);
        }
        if(u < 0) u += MD;
        return u;
    }

};
typedef ModInt<2015> mint;

int N;
struct Matrix {
    int a[maxn][maxn];

    Matrix() {
        for (int i = 0; i < N; i ++) {
            for (int j = 0; j < N; j ++) {
                a[i][j] = 0;
            }
        }
    }

    static Matrix unit() {
        Matrix ans;
        for (int i = 0; i < N; i ++) ans.a[i][i] = 1;
        return ans;
    }

    Matrix &operator * (const Matrix &that) const {
        static Matrix ans;
        for (int i = 0; i < N; i ++) {
            for (int j = 0; j < N; j ++) {
                ans.a[i][j] = 0;
                for (int k = 0; k < N; k ++) {
                    ans.a[i][j] += a[i][k] * that.a[k][j];
                    ans.a[i][j] %= 2015;
                }
            }
        }
        return ans;
    }

    static Matrix power(Matrix a, int n) {
        Matrix ans = unit(), buf = a;
        while (n) {
            if (n & 1) ans = ans * buf;
            buf = buf * buf;
            n >>= 1;
        }
        return ans;
    }
};

class Timer {
private:
    clock_t _start;
    clock_t _end;

public:
    void init() {
        _start = clock();
    }
    void get() {
        _end = clock();
        cout << double(_end - _start) / CLK_TCK << endl;
    }
};
Timer clk;


int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    int T, n, m, k, x;
    cin >> T;
    while (T --) {
        cin >> n >> m;
        Matrix a;
        N = n + 1;
        for (int i = 0; i < n; i ++) {
            scanf("%d", &k);
            for (int j = 0; j < k; j ++) {
                scanf("%d", &x);
                a.a[i][-- x] = 1;
            }
        }
        for (int i = 0; i < N; i ++) a.a[i][N - 1] = 1;
        Matrix A = Matrix::power(a, m - 1);
        mint ans = 0;
        for (int i = 0; i < N; i ++) {
            for (int j = 0; j < N; j ++) {
                ans += A.a[i][j];
            }
        }
        cout << ans.get() << endl;
    }
    return 0;
}

[hdu5411 CRB and Puzzle]DP,矩阵快速幂

标签:

原文地址:http://www.cnblogs.com/jklongint/p/4746679.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!