ZOJ 2853 Evolution 【简单矩阵快速幂】

这道题目第二次看的时候才彻底理解了是什么意思

把题目转化为数学模型分析后就是 有一个初始序列， 有一个进化率矩阵

求的是初始序列 与进化率矩阵进行 m 次运算后， 初始序列最后一位的答案

那么显然，可以对进化率矩阵进行快速幂计算

 

Example

Let's assume that P(0, 1) = P(1, 2) = 1, and at the beginning of a sub-process, the populations of 0, 1, 2 are 40, 20 and 10 respectively, then at the end of the sub-process, the populations are 0, 40 and 30 respectively.

 

这个栗子看懂了这题就会懂了。

 

 Source Code:

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler

#include <stdio.h>

#include <iostream>

#include <fstream>

#include <cstring>

#include <cmath>

#include <stack>

#include <string>

#include <map>

#include <set>

#include <list>

#include <queue>

#include <vector>

#include <algorithm>

#define Max(a,b) (((a) > (b)) ? (a) : (b))

#define Min(a,b) (((a) < (b)) ? (a) : (b))

#define Abs(x) (((x) > 0) ? (x) : (-(x)))

#define MOD 1000000007

#define pi acos(-1.0)



using namespace std;



typedef long long           ll      ;

typedef unsigned long long  ull     ;

typedef unsigned int        uint    ;

typedef unsigned char       uchar   ;



template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}

template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}



const double eps = 1e-7      ;

const int N = 210            ;

const int M = 1100011*2      ;

const ll P = 10000000097ll   ;



int n, m;



struct Mat{

    double mat[N][N];

};



Mat operator * (Mat a, Mat b){

    Mat c;

    memset(c.mat, 0, sizeof(c.mat));

    for(int k = 0; k < n; ++k){

        for(int i = 0; i < n; ++i){

            if(a.mat[i][k] <= 0)    continue;   //

            for(int j = 0; j < n; ++j){

                if(b.mat[k][j] <= 0)    continue;   //

                c.mat[i][j] += a.mat[i][k] * b.mat[k][j];

            }

        }

    }

    return c;

}



Mat operator ^ (Mat a, int k){

    Mat c;

    for(int i = 0; i < n; ++i){

        for(int j = 0; j < n; ++j){

            c.mat[i][j] = (i == j); //init

        }

    }

    for(; k; k >>= 1){

        if(k & 1)   c = c * a;  //key

        a = a * a;

    }

    return c;

}



int main(){

    int i, j, t, k, u, v, numCase = 0;

    while(EOF != scanf("%d%d",&n,&m)){

        if(0 == n && 0 == m)    break;

        double val;

        Mat a, b;

        memset(a.mat, 0, sizeof(a.mat));

        memset(b.mat, 0, sizeof(b.mat));

        for(i = 0; i < n; ++i)  b.mat[i][i] = 1;

        for(i = 0; i < n; ++i)  scanf("%lf", &a.mat[i][i]);

        scanf("%d",&t);

        while(t--){

            scanf("%d%d%lf",&u,&v,&val);

            b.mat[u][u] -= val; //

            b.mat[u][v] = val;  //

        }

        b = b ^ m;

        double cur = 0;

        for(i = 0; i < n; ++i){

            cur += b.mat[i][n - 1] * a.mat[i][i];

        }

        /*

        for(i = 0; i < n; ++i){

            for(j = 0; j < n; ++j){

                cout << c.mat[i][j] << ' ';

            }

            cout << endl;

        }

        */

        printf("%.0lf\n",cur);

    }

    return 0;

}



/*

3 1

40 20 10

2

0 1 1.0

1 2 1.0

*/