描述
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.
输入
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.
输出
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.
注意
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.
举例
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.
样例输入
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
样例输出
120
0
思路
第一关是读懂题意，我这种英语渣看了半天才看懂。
看懂题意后就很简单了，构建矩阵，然后标准的矩阵快速幂，注意最后取整就好。
顺便这里一个小插曲，我的devcpp报段错误，似乎是我把矩阵开太大的原因，但是数据范围就这么大，提交之后却AC了，也是尴尬。
代码
#include <cstdio>
#include<cstring>
#define ll long long
#define maxn 201
using namespace std;
int k;
int n;
struct Mat
{
double f[maxn][maxn];
void cls(){memset(f, 0, sizeof(f));}//全部置为0
Mat() {cls();}
friend Mat operator * (Mat a, Mat b)
{
Mat res;
for(int i = 0; i < n; i++) for(int j = 0; j < n; j++)
for(int k = 0; k < n; k++)
res.f[i][j] += a.f[i][k] * b.f[k][j];
return res;
}
};
Mat quick_pow(Mat a)
{
Mat ans;
for(int i = 0; i < n; i++) ans.f[i][i] = 1;
int b = k;
while(b != 0)
{
if(b & 1) ans = ans * a;
b >>= 1;
a = a * a;
}
return ans;
}
int main()
{
while(~scanf("%d %d", &n, &k))
{
if(n + k == 0) break;
Mat A, B;
for(int i = 0; i < n; i++) A.f[i][i] = 1;
for(int i = 0; i < n; i++)
scanf("%lf", &B.f[i][0]);
int t; scanf("%d", &t);
while(t--)
{
int x, y;
double p;
scanf("%d %d %lf", &x, &y, &p);
A.f[x][x] -= p;
A.f[y][x] += p;
}
A = quick_pow(A);
B = A * B;
printf("%.0lf
", B.f[n-1][0]);
}
return 0;
}