zoj 2853 Evolution

it2022-05-09 27

ZOJ Problem Set - 2853 Evolution

Time Limit: 5 Seconds Memory Limit: 32768 KB

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

120 0

Author: JIANG, Jiefeng

Source: Zhejiang Provincial Programming Contest 2007 Submit Status

// 1868165 2009-05-14 09:43:35 Wrong Answer 2853 C++ 3880 1472 Wpl // 1868191 2009-05-14 09:56:25 Wrong Answer 2853 C++ 3870 1472 Wpl // 1868298 2009-05-14 10:59:02 Accepted 2853 C++ 3900 1472 Wpl #include < iostream > #define MAX 203 using namespace std; typedef struct node { double matrix[MAX][MAX]; }Matrix; Matrix unit,init,result,c; int n,m,t; double data[MAX]; void Init() { int i,j; double p; for (i = 0 ;i < n;i ++ ) for (j = 0 ;j < n;j ++ ) { init.matrix[i][j] = (i == j); // 相当于init刚开始默认全部由自己进化到自己 unit.matrix[i][j] = (i == j); // 单位矩阵 } for (i = 0 ;i < n;i ++ ) scanf( " %lf " , & data[i]); scanf( " %d " , & t); while (t -- ) { scanf( " %d%d%lf " , & i, & j, & p); // init.matrix[i][j]+=p; // why wrong????? // init.matrix[i][i]-=p; init.matrix[j][i] += p; init.matrix[i][i] -= p; } } void Cal( int exp) { int i,j,k; if (exp == 1 ) result = init; while (exp != 1 ) { if (exp & 1 ) { exp -- ; for (i = 0 ;i < n;i ++ ) for (j = 0 ;j < n;j ++ ) { c.matrix[i][j] = 0 ; for (k = 0 ;k < n;k ++ ) c.matrix[i][j] += init.matrix[i][k] * unit.matrix[k][j]; } unit = c; } else { exp >>= 1 ; for (i = 0 ;i < n;i ++ ) for (j = 0 ;j < n;j ++ ) { c.matrix[i][j] = 0 ; for (k = 0 ;k < n;k ++ ) c.matrix[i][j] += init.matrix[i][k] * init.matrix[k][j]; } init = c; } } for (i = 0 ;i < n;i ++ ) for (j = 0 ;j < n;j ++ ) { result.matrix[i][j] = 0 ; for (k = 0 ;k < n;k ++ ) result.matrix[i][j] += unit.matrix[i][k] * init.matrix[k][j]; } } int main() { int i,j; double r; while (scanf( " %d%d " , & n, & m) != EOF) { if (n == 0 && m == 0 ) break ; Init(); Cal(m); // 求初始矩阵的m次幂 r = 0 ; for (j = 0 ;j < n;j ++ ) r += result.matrix[n - 1 ][j] * data[j]; printf( " %.0lf\n " ,r); } return 0 ; }

转载于:https://www.cnblogs.com/forever4444/archive/2009/05/14/1456595.html

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