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;    

 

/* ***********************************************
Author        :kuangbin
Created Time  :2013-11-17 23:18:47
File Name     :E:\2013ACM\比赛练习\2013-11-17\EE.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const double eps = 1e-9;
const int MAXN = 100;
double a[MAXN][MAXN],x[MAXN];
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(i = k+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(k != max_r)
        {
            for(j = col;j < var;j++)
                swap(a[k][j],a[max_r][j]);
            swap(x[k],x[max_r]);
        }
        x[k]/=a[k][col];
        for(j = col+1;j < var;j++)a[k][j]/=a[k][col];
        a[k][col] = 1;
        for(int i = 0;i < equ;i++)
            if(i != k)
            {
                x[i] -=  x[k]*a[i][k];
                for(j = col+1;j < var;j++)a[i][j] -= a[k][j]*a[i][col];
                a[i][col] = 0;
            }
    }
    return 1;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n,m;
    int T;
    int iCase = 0;
    scanf("%d",&T);
    while(T--)
    {
        iCase++;
        scanf("%d%d",&n,&m);
        equ = var = n;
        memset(a,0,sizeof(a));
        int u,v,w;
        for(int i = 0;i < m;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            a[u-1][v-1] += 1.0/w;
            a[u-1][u-1] += -1.0/w;
            a[v-1][u-1] += 1.0/w;
            a[v-1][v-1] += -1.0/w;
        }
        for(int i = 0;i < n-1;i++)
            x[i] = 0;
        x[0] = 1;
        for(int i = 0;i < n;i++)
            a[n-1][i] = 0;
        x[n-1] = 0;
        a[n-1][0] = 1;
        Gauss();
        printf("Case #%d: %.2lf\n",iCase,x[n-1]);
    }
    return 0;
}

 

 

 

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。