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Back to Kernighan-Ritchie
Input: Standard Input
Output: Standard Output
You must have heard the name of Kernighan and Ritchie, the authors of The C Programming Language. While coding in C, we use different control statements and loops, such as, if-then-else, for, do-while, etc. Consider the following fragment of pseudo code:
//execution starts here
do {
U;
V;
} while(condition);
W;
In the above code, there is a bias in each conditional branch. Such codes can be represented by control flow graphs like below:
Let the probability of jumping from one node of the graph to any of its adjacent nodes be equal. So, in the above code fragment, the expected number of times U executes is 2. In this problem, you will be given with such a control flow graph and find the expected number of times a node is visited starting from a specific node.
Input
Input consists of several test cases. There will be maximum 100 test cases. Each case starts with an integer: n (n ≤ 100). Here n is the number of nodes in the graph. Each node in the graph is labeled with 1 ton and execution always starts from 1. Each of the next few lines has two integers: start and end which means execution may jump from node startto node end. A value of zero for start ends this list. After this, there will be an integer q (q ≤ 100) denoting the number of queries to come. Next q lines contain a node number for which you have to evaluate the expected number of times the node is visited. The last test case has value of zero for n which should not be processed.
Output for each test case should start with ?Case #i:? with next q lines containing the results of the queries in the input with three decimal places. There can be situations where a node will be visited forever (for example, an infinite for loop). In such cases, you should print ?infinity? (without the quotes). See the sample output section for details of formatting.
|
3 1 2 2 3 2 1 0 0 3 1 2 3 3 1 2 2 3 3 1 0 0 3 3 2 1 0 |
Case #1: 2.000 2.000 1.000 Case #2: infinity infinity infinity |
Problem setter: Mohammad Sajjad Hossain
Special Thanks: Shahriar Manzoor
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 #include <cmath> 6 using namespace std; 7 8 const int maxn=110; 9 const double eps=1e-8; 10 typedef double Matrix[maxn][maxn]; 11 Matrix A; 12 int n,d[maxn];//d数组存i节点的初读 13 bool inf[maxn];//标记无穷变量 14 vector<int> pre[maxn];//存i节点的前驱 15 16 void swap(double &a,double &b){double t=a;a=b;b=t;} 17 18 void gauss_jordan() 19 { 20 int i,j,r,k; 21 for(i=0;i<n;i++) 22 { 23 r=i; 24 for(j=i+1;j<n;j++) 25 if(fabs(A[j][i])>fabs(A[r][i])) r=j; 26 if(fabs(A[r][i])<eps) continue; 27 if(r!=i)for(j=0;j<=n;j++) swap(A[r][j],A[i][j]); 28 //与第i行以外的其他行进行消元 29 for(k=0;k<n;k++) if(k!=i) 30 for(j=n;j>=i;j--) A[k][j]-=A[k][i]/A[i][i]*A[i][j]; 31 } 32 } 33 int main() 34 { 35 //freopen("in.txt","r",stdin); 36 //freopen("out.txt","w",stdout); 37 int i,j,icase=0; 38 while(scanf("%d",&n),n) 39 { 40 memset(d,0,sizeof(d)); 41 for(i=0;i<n;i++) pre[i].clear(); 42 int a,b; 43 while(scanf("%d %d",&a,&b),a) 44 { 45 a--;b--;d[a]++; 46 pre[b].push_back(a); 47 } 48 memset(A,0,sizeof(A)); 49 for(i=0;i<n;i++)//构造方程组 50 { 51 A[i][i]=1; 52 for(j=0;j<pre[i].size();j++) 53 A[i][pre[i][j]]-=1.0/d[pre[i][j]]; 54 if(i==0) A[i][n]=1; 55 } 56 //解方程组，标记无穷变量 57 gauss_jordan(); 58 memset(inf,0,sizeof(inf)); 59 for(i=n-1;i>=0;i--) 60 { 61 if(fabs(A[i][i])<eps && fabs(A[i][n])>eps) inf[i]=true;//这个变量无解，标记为无穷变量 62 for(j=i+1;j<n;j++)//跟无穷变量扯上关系的也是无穷的 63 if(fabs(A[i][j])>eps && inf[j]) inf[i]=true; 64 } 65 int q,p; 66 scanf("%d",&q); 67 printf("Case #%d:\n",++icase); 68 while(q--) 69 { 70 scanf("%d",&p);p--; 71 if(inf[p]) printf("infinity\n"); 72 else printf("%.3lf\n",fabs(A[p][p])<eps?0.0:A[p][n]/A[p][p]); 73 } 74 } 75 return 0; 76 }
uva 10828 高斯消元求数学期望,布布扣,bubuko.com
标签：des style blog color os strong
原文地址：http://www.cnblogs.com/xiong-/p/3861681.html
