關於 矩陣在ACM中的應用

關於矩陣在ACM中的應用

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一、矩陣運算法則

重點說說矩陣與矩陣的乘法，不說加減法。

支持：

• 結合律  (AB)C = A(BC)
• 分配律 A(B+C) = AB + AB
• $\left( \lambda A\right) B=\lambda \left( AB\right) =A\left( \lambda B\right) $

 

二、矩陣乘法的程序實現：

 

struct matrix { double a[200][200]; } ans, base; matrix multiply(matrix x, matrix y) { matrix tmp; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) { tmp.a[i][j] = 0; for (int k = 0; k < n; k++) tmp.a[i][j] += x.a[i][k] * y.a[k][j]; } return tmp; }

 

 

 

三、矩陣的冪運算能夠進行快速冪

　　由於矩陣的乘法支持結合律，因此B*A*A*A = B(A*A*A) = B*$A^{3}$

程序實現：

 

void fast_mod(int k)
{
    while (k)
    {
        if (k & 1) ans = multiply(ans, base);
        base = multiply(base, base);
        k >>= 1;
    }
}

 

 

 

 

四、將行列式的變換動做構形成一個矩陣（如 平移 旋轉 翻轉 等操做）

五、將一個一次遞推式 構造出 矩陣。

　　例如：fibonacci數列的遞推關係 f(n) = f(n-1) + f(n-2)

　　咱們能夠將矩陣構造爲：$\left( \begin{matrix} 0& 1\\ 1& 1\end{matrix} \right) \left( \begin{matrix} a\\ b\end{matrix} \right) =\left( \begin{matrix} b\\ a+b\end{matrix} \right) $

 

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好比今天作的一個題目  zoj 2853 Evolution

　　它就是先進行構造「進化」這個動做的矩陣。而後有多少次進化就等於"初始物種狀態矩陣" 乘 "進化矩陣」ⁿ 

 

題目 見文末

附上個人代碼：

#include<cstdio>
#include<cstring>
int n;
struct matrix
{
    double a[200][200];
} ans, base;

matrix multiply(matrix x, matrix y)
{
    matrix tmp;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
        {
            tmp.a[i][j] = 0;
            for (int k = 0; k < n; k++) tmp.a[i][j] += x.a[i][k] * y.a[k][j];
        }
    return tmp;
}
void fast_mod(int k)
{
    while (k)
    {
        if (k & 1) ans = multiply(ans, base);
        base = multiply(base, base);
        k >>= 1;
    }
}
int main()
{
    int m, t, xi, yi;
    double s, zi, fin;
    int num[205];
    while (~scanf("%d%d", &n, &m) && !(n == 0 && m == 0))
    {
        for (int i = 0; i < n; i++) scanf("%d", &num[i]);
        scanf("%d", &t);
        memset(base.a, 0, sizeof(base.a));
        for (int i = 0; i < n; i++) base.a[i][i] = 1;   //初始爲進化成本身
        while (t--)
        {
            scanf("%d%d%lf", &xi, &yi, &zi);
            base.a[xi][yi] += zi;
            base.a[xi][xi] -= zi;
        }
        memset(ans.a, 0, sizeof(ans.a));
        for (int i = 0; i < n; i++) ans.a[i][i] = 1;
        fast_mod(m);
        fin = 0;
        for (int i = 0; i < n; i++)
            fin += num[i] * ans.a[i][n-1];
        printf("%.0f\n", fin);
    }
    return 0;
}

View Code

 

六、矩陣在圖的鄰接矩陣中的應用

　　暫時還不會， = =、 之後補上。

 

 

 

 

 

Evolution

 

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Time Limit: 5 Seconds       Memory Limit: 32768 KB

 

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Evolution is a long, long process with extreme complexity and involves many species. Dr. C. P. Lottery is currently investigating a simplified model of evolution: consider that we have N (2 <= N <= 200) species in the whole process of evolution, indexed from 0 to N -1, and there is exactly one ultimate species indexed as N-1. In addition, Dr. Lottery divides the whole evolution process into M (2 <= M <= 100000) sub-processes. Dr. Lottery also gives an 'evolution rate' P(i, j) for 2 species i and j, where i and j are not the same, which means that in an evolution sub-process, P(i, j) of the population of species i will transform to species j, while the other part remains unchanged.

 

Given the initial population of all species, write a program for Dr. Lottery to determine the population of the ultimate species after the evolution process. Round your final result to an integer.

 

Input

 

The input contains multiple test cases!

 

Each test case begins with a line with two integers N, M. After that, there will be a line with N numbers, indicating the initial population of each species, then there will be a number T and T lines follow, each line is in format "i j P(i,j)" (0 <= P(i,j) <=1).

 

A line with N = 0 and M = 0 signals the end of the input, which should not be proceed.

 

Output

 

For each test case, output the rounded-to-integer population of the ultimate species after the whole evolution process. Write your answer to each test case in a single line.

 

Notes

 

• There will be no 'circle's in the evolution process.
• E.g. for each species i, there will never be a path i, s1, s2, ..., st, i, such that P(i,s1) <> 0, P(sx,sx+1) <> 0 and P(st, i) <> 0.
• The initial population of each species will not exceed 100,000,000.
• There're totally about 5 large (N >= 150) test cases in the input.

 

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.

 

Sample Input

 

2 3
100 20
1
0 1 1.0
4 100
1000 2000 3000 0
3
0 1 0.19
1 2 0.05
0 2 0.67
0 0

 

Sample Output

 

1200