Hdu5006 resistance (Gaussian elimination element)

Give you a complex network graph and tell you the equivalent resistance of S, T, and S and T. The method is to set the potential of S to 1, and the potential of t to 0. for each other vertex x, SIGMA (Ux-uy)/R (x, y) (that is, Y for each node connected to X, the potential difference is divided by the value of the resistance and 0, which should be the kihof's law or something). Then a bunch of equations are listed to solve the potential of each vertex. For all edges connected from The Source Vertex, find the current, know the total current, and know the total potential, you can know the resistance.

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <queue>#include <cassert>#include <vector>using namespace std;#define maxn 11000#define eps 1e-7struct Edge{    int u,v,c;    Edge(int ui,int vi,int ci):u(ui),v(vi),c(ci){}    Edge(){}};int dcmp(double x){    return (x>eps)-(x<-eps);}vector<Edge> E;vector<int> G[maxn];int n,m,s,t;int bel[maxn];int btot;void dfs(int u,int mark){    bel[u]=mark;    for(int i=0;i<G[u].size();++i){        int v=G[u][i];        if(bel[v]) continue;        else dfs(v,mark);    }}void getBelong(){    btot=0;    memset(bel,0,sizeof(bel));    for(int i=1;i<=n;++i){        if(!bel[i]){            ++btot;            dfs(i,btot);        }    }}int cnt[500][500];double mat[500][500];void gauss(int n){    for(int i=1;i<=n;++i){        int pivot=i;        for(int j=i;j<=n;++j){            if(abs(mat[j][i])>abs(mat[pivot][i])) pivot=j;        }        for(int x=1;x<=n+1;++x){            swap(mat[i][x],mat[pivot][x]);        }        if(abs(mat[i][i])<eps) continue;        for(int j=i+1;j<=n+1;++j){            mat[i][j]/=mat[i][i];        }        for(int j=1;j<=n;++j){            if(i!=j){                for(int k=i+1;k<=n+1;++k){                    mat[j][k]-=mat[j][i]*mat[i][k];                }            }        }    }}int main(){    int T;cin>>T;    while(T--){        scanf("%d%d%d%d",&n,&m,&s,&t);        for(int i=0;i<=n;++i) G[i].clear();        E.clear();        int ui,vi,ci;        for(int i=0;i<m;++i){            scanf("%d%d%d",&ui,&vi,&ci);            if(ci==0){                G[ui].push_back(vi);                G[vi].push_back(ui);            }            else{                E.push_back(Edge(ui,vi,ci));            }        }        getBelong();        if(bel[s]==bel[t]){            printf("%.6lf\n",0);continue;        }        memset(cnt,0,sizeof(cnt));        memset(mat,0,sizeof(mat));        for(int i=0;i<E.size();++i){            ui=E[i].u;vi=E[i].v;            if(bel[ui]!=bel[vi]){                cnt[bel[ui]][bel[vi]]++;                cnt[bel[vi]][bel[ui]]++;            }        }        for(int i=1;i<=btot;++i){            if(i==bel[s]){                mat[i][i]=1.0;                mat[i][btot+1]=1;                continue;            }            else if(i==bel[t]){                mat[i][i]=1.0;                mat[i][btot+1]=0;                continue;            }            for(int j=1;j<=btot;++j){                if(i==j||cnt[i][j]==0) continue;                mat[i][i]+=cnt[i][j];                mat[i][j]-=cnt[i][j];            }        }        gauss(btot);        double I=0;        for(int i=1;i<=btot;++i){            if(i==bel[s]) continue;            if(cnt[bel[s]][i]>0){                I+=(1-mat[i][btot+1])*cnt[bel[s]][i];            }        }        double R=1/I;        if(dcmp(I-eps)==0) {            printf("inf\n");            continue;        }        printf("%.6lf\n",R);    }    return 0;}

 

Hdu5006 resistance (Gaussian elimination element)