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矩阵快速幂。
构造一个矩阵，$a[i][j]$表示一次操作后，$j$会从$i$那里得到水的比例。注意$k=0$的时候，要将$a[i][j]$置为$1$。
#pragma comment(linker, "/STACK:1024000000,1024000000")#include #include #include #include #include #include #include #include #include #include #include using namespace std;typedef long long LL;const double pi=acos(-1.0);void File(){ freopen("D:\\in.txt","r",stdin); freopen("D:\\out.txt","w",stdout);}template inline void read(T &x){ char c = getchar(); x = 0; while(!isdigit(c)) c = getchar(); while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }}struct Matrix{ double A[25][25]; int R, C; Matrix operator*(Matrix b);};Matrix X, Y, Z;int T,n,p;Matrix Matrix::operator*(Matrix b){ Matrix c; memset(c.A, 0, sizeof(c.A)); int i, j, k; for (i = 1; i <= R; i++) for (j = 1; j <= b.C; j++) for (k = 1; k <= C; k++) c.A[i][j] = c.A[i][j] + A[i][k] * b.A[k][j]; c.R = R; c.C = b.C; return c;}void init(){ Y.R = n; Y.C = n; for (int i = 1; i <= n; i++) Y.A[i][i] = 1; X.R = n; X.C = n; Z.R = 1; Z.C = n;}void work(){ while (p) { if (p % 2 == 1) Y = Y*X; p = p >> 1; X = X*X; } Z = Z*Y;}int main(){ scanf("%d",&T); while(T--) { scanf("%d",&n); memset(X.A, 0, sizeof X.A); memset(Y.A, 0, sizeof Y.A); memset(Z.A, 0, sizeof Z.A); for(int i=1;i<=n;i++) scanf("%lf",&Z.A[1][i]); for(int i=1;i<=n;i++) { int K; scanf("%d",&K); for(int j=1;j<=K;j++) { int id; scanf("%d",&id); X.A[i][id]=1.0/K; } if(K==0) { X.A[i][i]=1.0; } } scanf("%d",&p); init(); work(); for(int i=1;i<=n;i++) { printf("%.2f",Z.A[1][i]); if(i
转载于:https://www.cnblogs.com/zufezzt/p/6407104.html