Electric resistance

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 907    Accepted Submission(s): 521

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

题意：给一个N个结点的电路图,任意两个结点均联通,

任意两点之间最多有一个电阻,求结点1与N之间的等效电阻。

分析：外加电流源求等效电阻,用结点电压法,以结点1为参考结点U₀=0,

即把结点1作为电势零点,结点2,3,4,...,N的电势为U_n1,U_n2,U_n3,...,U_n(N-1),

一共N-1个未知数,再在结点1与结点N之间加一个电流源I,列结点电压方程,

那么一共可列N-1个独立方程(含N-1个未知数),再用高斯消元解方程(一定有解),

解出U_n(N-1)，等效电阻R=(U_n(N-1)-U₀)/I=U_n(N-1)/I,为方便计算取I=1.0A，

那么R=U_n(N-1).

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
double a[60][60];
void Guass(int N,int M)
{
    for(int i=1;i<=N;i++)//i是列数,一个一个消去
    {//上三角
        if(fabs(a[i][i])<1e-6)//第i行第i个为0
        {//找出i+1到第M行中,第i个数不为0的行
            int p=M;
            for(;p>i;p--)
            if(fabs(a[p][i])>1e-6) break;
            if(p==i) continue;
            else swap(a[i],a[p]);//交换i,p两行
        }
        for(int k=i+1;k<=M;k++)
        {
            for(int j=i+1;j<=N+1;j++)
            {
                a[k][j]-=a[i][j]/a[i][i]*a[k][i];
            }
            a[k][i]=0;
        }
    }
    //从下往上消去
    for(int i=N;i;i--)//第i个解
    {
        for(int k=i-1;k;k--)//消去1到i-1行的第i个元素
        {
            a[k][N+1]-=a[i][N+1]/a[i][i]*a[k][i];
            a[k][i]=0;
        }
    }
}

int main()
{
    int T,cas=0;
    int N,M;//N个结点M条支路
    scanf("%d",&T);
    while(T--)
    {
        memset(a,0,sizeof(a));
        scanf("%d%d",&N,&M);
        for(int i=0;i<M;i++)
        {
            int s,t;double r;
            scanf("%d%d%lf",&s,&t,&r);
            //结点电压法
            if(s==1) a[t-1][t-1]+=1.0/r;
            else if(t==1) a[s-1][s-1]+=1.0/r;
            else
            {
                a[s-1][s-1]+=1.0/r;a[s-1][t-1]-=1.0/r;
                a[t-1][t-1]+=1.0/r;a[t-1][s-1]-=1.0/r;
            }
        }
        a[N-1][N]=1.0;//外加电流源
        Guass(N-1,N-1);//N-1个未知数
        printf("Case #%d: %.2lf\n",++cas,fabs(a[N-1][N]/a[N-1][N-1]));
    }
    return 0;
}

原文地址：https://www.cnblogs.com/ACRykl/p/8711227.html