2015
04-14

# Electric resistance

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.

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!

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!

1
4 5
1 2 1
2 4 4
1 3 8
3 4 19
2 3 12

Case #1: 4.21

#include <CSTDIO>
#include <QUEUE>
using namespace std;

/*
hdu 3976

R = 总电势差/电流

u[i]代表节点i的电势
Σ((u[j]-u[i])/r[i][j]) + I = 0
I[0] = 1
I[n-1] = -1
*/
const int MAXN = 55;
const double _inf = 1e-9;
double a[MAXN][MAXN], x[MAXN]; // 方程左边的矩阵和等式右边的值， x存放最后结果
int equ, val;	// 方程数 未知数个数
inline double mabs(double _X){return _X<0?-_X:_X;}
int gauss()
{
int i,j,k,col,max_r;
for(k=0,col=0;k<equ&&col<val;k++,col++)
{
max_r=k;
for(i=k+1;i<equ;i++)
{
if(mabs(a[i][col])>mabs(a[max_r][col]))
max_r=i;
}
if(mabs(a[max_r][col])<_inf) return 0;
if(k!=max_r)
{
for(j=col;j<val;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<val;j++)a[k][j]/=a[k][col];
a[k][col]=1;
for(i=0;i<equ;i++)
{
if(i!=k)
{
x[i]-=x[k]*a[i][k];
for(j=col+1;j<val;j++)a[i][j]-=a[k][j]*a[i][col];
a[i][col]=0;
}
}
}
return 1;
}

int n, m;
double data[MAXN][MAXN];
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif
int t, cs = 0;
int i, j;
double tmp;
scanf("%d", &t);
while (t--)
{
scanf("%d%d", &n, &m);
memset(data, 0, sizeof data);
while (m--)
{
scanf("%d%d%lf", &i, &j, &tmp);
--i, --j;
data[i][j] = data[j][i] = tmp;
}
memset(a, 0, sizeof a);
memset(x, 0, sizeof x);
for (i = 0; i< n; ++i)
{
for (j = 0; j< n; ++j)
{
if (data[i][j])
{
a[i][j] += 1.0/data[i][j];
a[i][i] -= 1.0/data[i][j];
}
}
}
x[0] = -1;
x[n-1] = 1;
equ = val = n;
gauss();
printf("Case #%d: %.2lf\n", ++cs, x[0]-x[n-1]);
}
return 0;
}


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2. 这道题这里的解法最坏情况似乎应该是指数的。回溯的时候
O(n) = O(n-1) + O(n-2) + ….
O(n-1) = O(n-2) + O(n-3)+ …
O(n) – O(n-1) = O(n-1)
O(n) = 2O(n-1)