ZOJ 2974-Just Pour the Water（矩阵快速幂）

Description

Shirly is a very clever girl. Now she has two containers (A and B), each with some water. Every minute, she pours half of the water in A into B, and simultaneous pours half of the water in B into A. As the pouring continues, she finds it is very easy to calculate the amount of water in A and B at any time. It is really an easy job :).

But now Shirly wants to know how to calculate the amount of water in each container if there are more than two containers. Then the problem becomes challenging.

Now Shirly has N (2 <= N <= 20) containers (numbered from 1 toN). Every minute, each container is supposed to pour water into anotherK containers (K may vary for different containers). Then the water will be evenly divided intoK portions and accordingly poured into antherK containers. Now the question is: how much water exists in each container at some specified time?

For example, container 1 is specified to pour its water into container 1, 2, 3. Then in every minute, container 1 will pour its 1/3 of its water into container 1, 2, 3 separately (actually, 1/3 is poured back to itself, this is allowed by the rule of the game).

Input

Standard input will contain multiple test cases. The first line of the input is a single integerT (1 <=T <= 10) which is the number of test cases. And it will be followed byT consecutive test cases.

Each test case starts with a line containing an integer N, the number of containers. The second line containsN floating numbers, denoting the initial water in each container. The followingN lines describe the relations that one container(from 1 to N) will pour water into the others. Each line starts with an integerK (0 <=K <= N) followed by K integers. Each integer ([1,N]) represents a container that should pour water into by the current container. The last line is an integerM (1<=M <= 1,000,000,000) denoting the pouring will continue forM minutes.

Output

For each test case, output contains N floating numbers to two decimal places, the amount of water remaining in each container after the pouring in one line separated by one space. There is no space at the end of the line.

Sample Input

12100.00 100.001 22 1 22

Sample Output

75.00 125.00

Note: the capacity of the container is not limited and all the pouring at every minute is processed at the same time.

                                                                                  

理解了好久的题意>.<...有N 杯水，N种操作，操作次数为M;

思路：f1(n）指的是第一杯水第n 次操作得到的水的数量~

            根据样例，f1(1)=0*f1(0)+1/2*f2(0)；f2(1)=1*f1(0)+1/2*f2(0);

                               f1(2)=0*f1(1)+1/2*f2(1)；f2(2)=1*f1(1)+1/2*f2(1);

           由递推式可以发现可以得到一个矩阵，矩阵中数是方程的系数   0   1/2

                                                                                                                         1/2 1/2

            矩阵快速幂~~

CODE：

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <string>#include <cstring>#include <queue>#include <stack>#include <vector>#include <set>#include <map>const int inf=0xfffffff;typedef long long ll;using namespace std;typedef vector<double> vec;typedef vector<vec> mat;mat mul(mat &A,mat &B){    mat C(A.size(),vec(B[0].size()));    for(int i=0;i<A.size();i++){        for(int k=0;k<B.size();k++){            for(int j=0;j<B[0].size();j++){                C[i][j]=C[i][j]+A[i][k]*B[k][j];            }        }    }    return C;}mat pow(mat A,ll m){    mat B(A.size(),vec(A.size()));    for(int i=0;i<B.size();i++)        B[i][i]=1.0;    while(m>0){        if(m & 1) B=mul(B,A);        A=mul(A,A);        m>>=1;    }    return B;}int main(){    //freopen("in.in","r",stdin);    int T,N;    scanf("%d",&T);    while(T--){        double n[25];        scanf("%d",&N);        for(int i=0;i<N;i++){            scanf("%lf",&n[i]);        }        mat A(N,vec(N));        int k,a;        for(int i=0;i<N;i++){//第i个容器            scanf("%d",&k);            if(k==0) A[i][i]=1.0;            for(int j=0;j<k;j++){                scanf("%d",&a);                A[a-1][i]+=(1.0/k);            }        }        ll M;        scanf("%lld",&M);        A=pow(A,M);        for(int i=0;i<N;i++){            double p=0;            for(int j=0;j<N;j++){                p+=A[i][j]*n[j];            }            if(i==0) printf("%.2f",p);            else printf(" %.2f",p);        }        printf("\n");    }    return 0;}

                              f2(1)=1*f1(0)+1/2*f2(0);