HDU 3976 Electric resistance （高斯消元法）

Electric resistance

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 326    Accepted Submission(s): 156

Problem 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

 

 

Author

abcdxyzk

 

 

Source

2011 Multi-University Training Contest 14 - Host by FZU

 

 

 

高斯消元解方程组。

 

主要是方程的建立。

 

我建方程使用了n个未知数，表示n个点的电势。

 

需要列n个方程。

 

就根据n个点，流入电流等于流出电流，或者说每个点电流之和（假如流入为正，流出为负，反之也可）

这样可以列出n个方程，根据n个点电流和为0.

 

而且可以假设1这个点流入电流为-1, 这样设点电势为0，那么可以知道n这个点的电势就等于等效电阻了、。

 

流入肯定等于流出的，上面列的方程组中第n个的是多余的，可以去掉，替换成1点电压为0.

这样方程组正确建立。

 

对于u  ---->  v  电阻为w.   可以知道u加一个电流  xv/w - xu/w.  而v加一个电流 xu/w - xv/w;    

 

 1 /* ***********************************************

 2 Author        :kuangbin

 3 Created Time  :2013-11-17 23:18:47

 4 File Name     :E:\2013ACM\比赛练习\2013-11-17\EE.cpp

 5 ************************************************ */

 6 

 7 #include <stdio.h>

 8 #include <string.h>

 9 #include <iostream>

10 #include <algorithm>

11 #include <vector>

12 #include <queue>

13 #include <set>

14 #include <map>

15 #include <string>

16 #include <math.h>

17 #include <stdlib.h>

18 #include <time.h>

19 using namespace std;

20 const double eps = 1e-9;

21 const int MAXN = 100;

22 double a[MAXN][MAXN],x[MAXN];

23 int equ,var;

24 int Gauss()

25 {

26     int i,j,k,col,max_r;

27     for(k = 0,col = 0;k < equ && col < var;k++,col++)

28     {

29         max_r = k;

30         for(i = k+1;i < equ;i++)

31             if(fabs(a[i][col]) > fabs(a[max_r][col]))

32                 max_r = i;

33         if(fabs(a[max_r][col]) < eps)return 0;

34         if(k != max_r)

35         {

36             for(j = col;j < var;j++)

37                 swap(a[k][j],a[max_r][j]);

38             swap(x[k],x[max_r]);

39         }

40         x[k]/=a[k][col];

41         for(j = col+1;j < var;j++)a[k][j]/=a[k][col];

42         a[k][col] = 1;

43         for(int i = 0;i < equ;i++)

44             if(i != k)

45             {

46                 x[i] -=  x[k]*a[i][k];

47                 for(j = col+1;j < var;j++)a[i][j] -= a[k][j]*a[i][col];

48                 a[i][col] = 0;

49             }

50     }

51     return 1;

52 }

53 int main()

54 {

55     //freopen("in.txt","r",stdin);

56     //freopen("out.txt","w",stdout);

57     int n,m;

58     int T;

59     int iCase = 0;

60     scanf("%d",&T);

61     while(T--)

62     {

63         iCase++;

64         scanf("%d%d",&n,&m);

65         equ = var = n;

66         memset(a,0,sizeof(a));

67         int u,v,w;

68         for(int i = 0;i < m;i++)

69         {

70             scanf("%d%d%d",&u,&v,&w);

71             a[u-1][v-1] += 1.0/w;

72             a[u-1][u-1] += -1.0/w;

73             a[v-1][u-1] += 1.0/w;

74             a[v-1][v-1] += -1.0/w;

75         }

76         for(int i = 0;i < n-1;i++)

77             x[i] = 0;

78         x[0] = 1;

79         for(int i = 0;i < n;i++)

80             a[n-1][i] = 0;

81         x[n-1] = 0;

82         a[n-1][0] = 1;

83         Gauss();

84         printf("Case #%d: %.2lf\n",iCase,x[n-1]);

85     }

86     return 0;

87 }

 

 

 

第一次写的时候用n+m个未知数做的，也可以A掉，但是有m个变量多余了。