hdu3976--Electric resistance(高斯消元问题7)

Electric resistance

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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Description

Now give you a circuit who has n nodes (marked from 1 to n) , please tell abcdxyzk the equivalent resistance of the circuit between node 1 and node n. You may assume that the circuit is connected. The equivalent resistance of the circuit between 1 and n is that, if you only consider node 1 as positive pole and node n as cathode , all the circuit could be regard as one resistance . (It's important to analyse complicated circuit ) At most one resistance will between any two nodes.  

 

Input

In the first line has one integer T indicates the number of test cases. (T <= 100)  

Each test first line contain two number n m(1<n<=50,0<m<=2000), n is the number of nodes, m is the number of resistances.Then follow m lines ,each line contains three integers a b c, which means there is one resistance between node a and node b whose resistance is c. (1 <= a,b<= n, 1<=c<=10^4) You may assume that any two nodes are connected!

 

Output

for each test output one line, print "Case #idx: " first where idx is the case number start from 1, the the equivalent resistance of the circuit between 1 and n. Please output the answer for 2 digital after the decimal point .

 

Sample Input

      
      
      
      

       
       
       
        1 
4 5 
1 2 1 
2 4 4 
1 3 8 
3 4 19 
2 3 12 
      
      
      
      

 

Sample Output

      
      
      
      

       
       
       
        Case #1: 4.21 
      
      
      
      

给出n个节点，和每个节点中的电阻，问总的电阻是多少。

设每个节点的电势为ui那么总的电压为un - u1 ， 设流过这个电路的电流为1，那么电阻也就是un - u1了。通过每个节点流入的电流和流出的电流相同，得到n个方程。

设在1节点流入的是-1，n节点流出的是1，那么对于u到v有电阻w，也就是u点的电流 (v-u)/w，对于v点的电流为（u-v）/w ;

得到n个方程后高斯消元，设1节点的电势为0，这样就能直接求出n节点的电势，也就是整个的电阻了。

 

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std ;
#define eps 1e-8
double Map[60][60] , a[60] , x[60] ;
void swap1(int p,int q,int n)
{
    int i ;
    double temp ;
    temp = a[p] ; a[p] = a[q] ; a[q] = temp ;
    for(i = 1;  i <= n ; i++)
    {
        temp = Map[p][i] ; Map[p][i] = Map[q][i] ; Map[q][i] = temp ;
    }
    return ;
}
double gauss(int n)
{
    int i , j , k , t , max1 ;
    double temp ;
    for(i = 1 , t = 1 ; i <= n && t <= n ; i++ , t++)
    {
        max1 = i ;
        for(j = i ; j <= n ; j++)
            if( fabs(Map[j][t])-fabs(Map[max1][t]) > 0 )
            {
                max1 = j ;
            }
        if( fabs(Map[max1][t]) < eps )
        {
            i-- ;
            continue ;
        }
        if( i != max1 )
            swap1(i,max1,n) ;
        for(j = i+1 ; j <= n ; j++)
        {
            if( fabs(Map[j][t]) < eps ) continue ;
            temp = Map[j][t] / Map[i][t] ;
            a[j] -= (a[i]*temp) ;
            for(k = t ; k <= n ; k++)
                Map[j][k] -= (Map[i][k]*temp) ;
        }
    }
    for(i = n ; i >= 1 ; i--)
    {
        x[i] = a[i] ;
        for(j = i+1 ; j <= n ; j++)
            x[i] -= ( Map[i][j]*x[j] ) ;
        x[i] /= Map[i][i] ;
    }
    return fabs(a[n-1]/Map[n-1][n]) ;
}
int main()
{
    int t , tt , n , m , u , v ;
    double w ;
    scanf("%d", &t) ;
    for(tt = 1 ; tt <= t ; tt++)
    {
        memset(Map,0,sizeof(Map)) ;
        memset(a,0,sizeof(a)) ;
        scanf("%d %d", &n, &m) ;
        a[1] = -1.0 ;
        a[n] = 1.0 ;
        while(m--)
        {
            scanf("%d %d %lf", &u, &v, &w) ;
            Map[u][u] -= 1.0/w ;
            Map[u][v] += 1.0/w ;
            Map[v][u] += 1.0/w ;
            Map[v][v] -= 1.0/w ;
        }
        for(int i = 1 ; i <= n ; i++)
            Map[i][1] = 0 ;
        printf("Case #%d: %.2lf\n", tt, gauss(n) );
    }
    return 0;
}