HDU 3976 Electric resistance(高斯消元)

一看图就233了啊，电路这种高大上的东东，简直了啊。

这是binshen写的，看的他的博客：http://www.cnblogs.com/kuangbin/p/3428573.html

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;    

然后就找到了关系方程组。

可以边化简，边解方程。

Electric resistance

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 377    Accepted Submission(s): 183


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

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-10
///#define M 1000100
#define LL __int64
///#define LL long long
#define INF 0x7ffffff
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 2

const int maxn = 210;

using namespace std;

double a[maxn][maxn];
double x[maxn];
int equ, var;
int Gauss()
{
int row, col;
int max_r;
row = col = 0;
while(row < equ && col < var)
{
max_r = row;
for(int i = row+1; i < equ; i++)
if(fabs(a[i][col]) > fabs(a[max_r][col])) max_r = i;
if(fabs(a[max_r][col]) < eps) return 0;
if(row != max_r)
{
for(int j = col; j < var; j++) swap(a[row][j], a[max_r][j]);
swap(x[max_r], x[row]);
}
x[row] /= a[row][col];
for(int j = col+1; j < var; j++) a[row][j] /= a[row][col];
a[row][col] = 1;
for(int i = row+1; i < equ; i++)
{
x[i] -= x[row]*a[i][row];
for(int j = col+1; j < var; j++) a[i][j] -= a[row][j]*a[i][col];
a[i][col] = 0;
}
col++;
row++;
}
return 1;
}

int main()
{
int T;
int Case = 1;
cin >>T;
while(T--)
{
int n, m;
cin >>n>>m;
memset(a, 0, sizeof(a));
memset(x, 0, sizeof(x));
int u, v, w;
for(int i = 0; i < m; i++)
{
cin >>u>>v>>w;
a[u-1][u-1] += -1.0/w;
a[v-1][v-1] += -1.0/w;
a[u-1][v-1] += 1.0/w;
a[v-1][u-1] += 1.0/w;
}
x[0] = 1;
for(int i = 0; i < n; i++) a[n-1][i] = 0;
x[n-1] = 0;
a[n-1][0] = 1;
equ = var = n;
Gauss();
cout<<"Case #"<<Case++<<": ";
printf("%.2lf\n",x[n-1]);
}
return 0;
}