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!
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Un1,Un2,Un3,...,Un(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Un(N-1)，等效电阻R=(Un(N-1)-U0)/I=Un(N-1)/I,为方便计算取I=1.0A，
那么R=Un(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; }
