HDU 3976 (高斯消元 KCL方程)

Electric resistance

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 498    Accepted Submission(s): 259

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

 

真是个神题啊，给跪了～～深深地感受到了来自电路原理的恶意。

题意是给出电路上的n个点，然后m个电阻，告诉你每个电阻所在的左右两个点，求所有电阻的等效电阻。

根据基尔霍夫定律，流入或流出某一节点的所有之路的电流代数和为0。我们假设这些点是0-n-1并且0这个点的电势为0，那么只需要求出n-1这个点的电势。假设流入1的总电流是1A，那么等效电阻在数值上等于n-1这个点的电势。

然后根据基尔霍夫定律对每一个电阻更新KCL方程，x[i]表示i这个点的电势。

最后注意第一个点的电势是已知的，假设是0。因为我们无法求出每个点的绝对电势。如果忘记这一点KCL方程会跑出很奇怪的答案，因为解没有唯一性。

#include <bits/stdc++.h>using namespace std;#define maxn 111#define eps 1e-9double a[maxn][maxn], x[maxn];int n, m;int equ, var;int Gauss () {    int i, j, k, col, max_r;    for (k = 0, col = 0; k < equ && col < var; k++, col++) {        max_r = k;        for (int i = k+1; i < equ; i++) { //找到最大的数所在的行            if (fabs (a[i][col]) > fabs (a[max_r][col]))                max_r = i;        }        if (k != max_r) { //交换行            for (int j = col; j <= var; j++) {                swap (a[k][j], a[max_r][j]);            }        }        for (int i = k+1; i < equ; i++) { //消去            if (a[i][col]) {                double tmp = -a[i][col]/a[k][col];                for (int j = col; j <= var; j++) {                    a[i][j] += tmp*a[k][j];                }            }        }    }    for (int i = equ-1; i >= 0; i--) { //回代        double tmp = a[i][var];        for (int j = i+1; j < var; j++) {            tmp -= x[j]*a[i][j];        }        x[i] = tmp/a[i][i];    }    return 1;}int main () {    //freopen ("in", "r", stdin);    int kase = 0, t;    cin >> t;    while (t--) {        cin >> n >> m;        memset (a, 0, sizeof a);        equ = n, var = n;        a[0][var] = 1, a[n-1][var] = -1;        for (int i = 1; i <= m; i++) {            int u, v;            double r;            cin >> u >> v >>r; if (v < u) swap (u, v);            u--, v--;            a[u][u] -= 1.0/r;            a[u][v] += 1.0/r;            a[v][v] -= 1.0/r;            a[v][u] += 1.0/r;        }        for (int i = 0; i < n; i++) {            a[i][0] = 0;        }        Gauss ();        printf ("Case #%d: %.2f\n", ++kase, x[n-1]);    }    return 0;}