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-      ;
const int N =             ;
const int M = *      ;
const ll P = 10000000097ll   ;

int n, m;

struct Mat{
    double mat[N][N];
};

Mat operator * (Mat a, Mat b){
    Mat c;
    memset(c.mat, , sizeof(c.mat));
    for(int k = ; k < n; ++k){
        for(int i = ; i < n; ++i){
            if(a.mat[i][k] <= )    continue;   //
            for(int j = ; j < n; ++j){
                if(b.mat[k][j] <= )    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 = ; i < n; ++i){
        for(int j = ; j < n; ++j){
            c.mat[i][j] = (i == j); //init
        }
    }
    for(; k; k >>= ){
        if(k & )   c = c * a;  //key
        a = a * a;
    }
    return c;
}

int main(){
    int i, j, t, k, u, v, numCase = ;
    while(EOF != scanf("%d%d",&n,&m)){
        if( == n &&  == m)    break;
        double val;
        Mat a, b;
        memset(a.mat, , sizeof(a.mat));
        memset(b.mat, , sizeof(b.mat));
        for(i = ; i < n; ++i)  b.mat[i][i] = ;
        for(i = ; 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 = ;
        for(i = ; i < n; ++i){
            cur += b.mat[i][n - ] * 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 ;
}

/*
3 1
40 20 10
2
0 1 1.0
1 2 1.0
*/