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[UVA][高斯消去、期望值] 10828 - Back to Kernighan-Ritchie

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Problem I
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 to n 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 start to 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

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.

 

Sample Input                                  Output for Sample Input

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


由於期望值 E[j] = sigma(E[i]*(1/deg[i])), if i->j exists.
將每條方程式展開, 因此會有 n 條方程式, n 個一元的未知數求解。
由於起始點的期望值比較特別
E[1] = sigma(E[i]*(1/deg[i]))+1
大致上就是如此,利用高斯消去法求解每個方程式的解。

至於卡在一點,算方程式的解時,由於分子太小,double 跑到 NaN 去了。
特別判斷一下這個例子。

#include <stdio.h>
#include <vector>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
double sol[105];
int nosol[105];
void gaussianElimination(double mtx[][105], int n) {
#define eps 1e-8
    int i, j;
    for(i = 1; i <= n; i++) {
        int k = i;
        for(j = i+1; j <= n; j++)
            if(fabs(mtx[k][i]) < fabs(mtx[j][i]))
                k = j;
        if(fabs(mtx[k][i]) < eps)
            continue;
        if(k != i) {
            for(j = 1; j <= n+1; j++)
                swap(mtx[k][j], mtx[i][j]);
        }
        for(j = 1; j <= n; j++) {
            if(i == j)  continue;
            for(k = n+1; k >= i; k--) {
                mtx[j][k] -= mtx[j][i]/mtx[i][i]*mtx[i][k];
            }
        }
    }
    memset(nosol, 0, sizeof(nosol));
    for(i = n; i >= 1; i--) {
        if(fabs(mtx[i][n+1]) > eps && fabs(mtx[i][i]) < eps)
            nosol[i] = 1;
        else {
            if(fabs(mtx[i][n+1]) < eps)
                sol[i] = 0;
            else
                sol[i] = mtx[i][n+1]/mtx[i][i];
        }
        for(j = i+1; j <= n; j++)
            if(fabs(mtx[i][j]) > eps && nosol[j])
                nosol[i] = 1;
    }
}
int main() {
    int cases = 0;
    int n, x, y, i, j;
    while(scanf("%d", &n) == 1 && n) {
        vector<int> invg[105];
        int deg[105] = {};
        while(scanf("%d %d", &x, &y) == 2) {
            if(x == 0 && y == 0)
                break;
            deg[x]++;
            invg[y].push_back(x);
        }
        double mtx[105][105];
        memset(mtx, 0, sizeof(mtx));
        mtx[1][n+1] = 1;
        for(i = 1; i <= n; i++) {
            mtx[i][i] = 1;
            for(j = 0; j < invg[i].size(); j++)
                mtx[i][invg[i][j]] -= 1.0/deg[invg[i][j]];
        }
        gaussianElimination(mtx, n);
        printf("Case #%d:\n", ++cases);
        scanf("%d", &x);
        while(x--) {
            scanf("%d", &y);
            if(nosol[y])
                puts("infinity");
            else
                printf("%.3lf\n", sol[y]);
        }
    }
    return 0;
}
/*
E[j] = sigma(E[i]*(1/deg[i])), if i->j exists.
*/

台長: Morris
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