[矩阵幂运算] ZOJ 2974 Just Pour the Water

理解题意之后，这题就是一个简单地矩阵幂运算，用类似二分的方法水过~~

值得注意的是，当k为0时，水将保留在原来的瓶子里。

 

#include<stdio.h>#include<string.h>const int maxn = 21;struct Matrix{double matrix[maxn][maxn];};double ans[maxn];Matrix init, unit;int n;Matrix matrix_mul(Matrix a, Matrix b){Matrix c;int i, j, kk;for(i = 0; i < n; i++)for(j = 0; j < n; j++){c.matrix[i][j] = 0;for(kk = 0; kk < n; kk++)c.matrix[i][j] += a.matrix[i][kk]*b.matrix[kk][j];}return c;}Matrix matrix_pow(Matrix a, int m){if(m == 1)return a;else if(m & 1)return matrix_mul(a, matrix_pow(a, m-1));else{Matrix b = matrix_pow(a, m/2);return matrix_mul(b, b);}/*Matrix i = init, u = unit;while(m != 1){if(m%2 != 0){--m;u = matrix_mul(i, u);}else{m /= 2;i = matrix_mul(i, i);}}i = matrix_mul(i, u);return i;*/}//#define LOCALint main(){#ifdef LOCAL freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);#endifint test;int i, j, k, m;scanf("%d", &test);for(int i = 0; i < maxn; i++)unit.matrix[i][i] = 1;while(test--){memset(init.matrix, 0.0, sizeof(init.matrix));scanf("%d", &n);for( i = 0; i < n; i++)scanf("%lf", &ans[i]);for( i = 0; i < n; i++){scanf("%d", &k);if(k == 0)init.matrix[i][i] = 1.0;else{double tmp = 1.0/k;for( j = 0; j < k; j++){scanf("%d", &m);init.matrix[m-1][i] = tmp;}}}scanf("%d", &m);Matrix temp = matrix_pow(init, m);double res;for(int i = 0; i < n-1; i++){res = 0.0;for(int j = 0; j < n; j++)res += temp.matrix[i][j] * ans[j];printf("%.2lf ", res);}res = 0.0;for(int j = 0; j < n; j++)res += temp.matrix[n-1][j] * ans[j];printf("%.2lf/n", res);}return 0;}