【BZOJ 2419】电阻 高斯消元

%物理大神

重点是设电势为未知量

#include<cstdio>#include<cstring>#include<iostream>#define maxn 151using namespace std;int n,m,head[maxn],tot=1;struct edge{int v,next;double w;}e[maxn*maxn*2];void adde(int a,int b,double c){e[tot].v=b,e[tot].next=head[a],e[tot].w=c;head[a]=tot++;}double a[maxn][maxn];void solve(){for(int p,i=1;i<=n;i++){for(p=i;p<=n;p++)if(a[p][i])break;if(p>n)continue;for(int j=1;j<=n+1;j++)swap(a[i][j],a[p][j]);double z=a[i][i];for(int j=1;j<=n+1;j++)a[i][j]/=z;for(int j=1;j<=n;j++)if(a[j][i]&&j!=i){double d=a[j][i]/a[i][i];for(int k=1;k<=n+1;k++)a[j][k]-=a[i][k]*d;}}}int main(){while(scanf("%d%d",&n,&m)!=EOF){tot=1;memset(head,0,sizeof(head));memset(a,0,sizeof(a));int x,y;double c;for(int i=1;i<=m;i++){scanf("%d%d%lf",&x,&y,&c);adde(x,y,c),adde(y,x,c);}for(int i=1;i<n;i++){for(int j=head[i];j;j=e[j].next){int v=e[j].v;a[i][v]+=1.0/e[j].w;a[i][i]-=1.0/e[j].w;}}a[1][n+1]=-1;for(int i=1;i<=n+1;i++)a[n][i]=0;a[n][n]=1;solve();printf("%.2lf\n",a[1][n+1]);}return 0;}