集训队专题（2）1007 Evolution

Evolution

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 52   Accepted Submission(s) : 18

Problem 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

 

Source

Zhejiang Provincial Programming Contest 2007

 

此题也是比较基础的矩阵快速幂，注意浮点数的类型就好了

#include <cstdio>#include <cstring>#define MAX 203struct Matrix{double mat[MAX][MAX];}init,unit;double pop[MAX];int N,M,T;void Init(){memset(init.mat,0,sizeof(init.mat));memset(unit.mat,0,sizeof(unit.mat));for(int i=0; i<N; i++){init.mat[i][i] = 1;//开始默认全部由自己进化到自己unit.mat[i][i] = 1;//初始为单位矩阵}for(int i=0; i<N; i++)scanf("%lf",&pop[i]);scanf("%d",&T);int i,j;double p;while(T--){scanf("%d%d%lf",&i,&j,&p);init.mat[j][i] += p;init.mat[i][i] -= p;}}Matrix multi(Matrix a,Matrix b){Matrix c;for(int i=0; i<N; i++){for(int j=0; j<N; j++){c.mat[i][j] = 0;for(int k=0; k<N; k++){c.mat[i][j] += a.mat[i][k]*b.mat[k][j];}}}return c;}Matrix fast_pow(int x){Matrix i=init,ans=unit;      while(x){          if(x&1)              ans = multi(ans,i);          i = multi(i,i);          x >>= 1;    }      return ans;}int main(){double ans;while(scanf("%d%d",&N,&M)&&(N||M)){Init();Matrix result=fast_pow(M);ans = 0;for(int i=0; i<N; i++){ans += result.mat[N-1][i]*pop[i];}printf("%.0lf\n",ans);}return 0;}