傳送門：ZOJ 1853

Description

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

題目大意：總體是一個進化的過程，共有n個物種，p(i,j)表示一輪進化中i->j轉化的比例，問裝化m輪后第n個物種的個數。

思路：因為每輪都是對整個物種的數組進行乘法，然后加起來，正好用矩陣乘法。然后用快速冪加速。

代碼：

#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib>using namespace std;const int N=250; int n; 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; //優化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); //初始化為單位矩陣}}while(k){if(k&1) c=c*a;a=a*a;k>>=1;}return c;} int main() {int m;//freopen("in.txt","r",stdin);while(scanf("%d%d",&n,&m),n||m){Mat species,p;for(int i=0;i<n;i++)for(int j=0;j<n;j++)p.mat[i][j]=(i==j);for(int i=0;i<n;i++)scanf("%lf",&species.mat[i][0]);int nn;scanf("%d",&nn);int j,k;double v;for(int i=0;i<nn;i++){cin>>j>>k>>v;p.mat[k][j]+=v;p.mat[j][j]-=v;}p=p^m;species=p*species;printf("%.0f\n",species.mat[n-1][0]);//cout<<"----"<<endl;}return 0; }

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