HDU 5006 Resistance 物理 高斯消元

题意：给出N个点，M条导线，导线的电阻为0或1，求出给定两点的等效电阻。

思路：电路题呀。全忘光了。。。。

           首先，用电阻为0的导线连成的两点是等电势的。这样，我们将这些点可以缩成一个点。

           等效电阻为R = U  /  I.当给定电流I = 1A时，给定两点的等效电阻就是两点的电势之差。

           然后。。。然后就是节点电压法了。将缩点后的电路图，求出自电导和互电导，然后带入方程组。同时假设任意一点为参考点，高斯消元求出解即可。

代码如下：

#include <cstdio>#include <cstring>#include <cmath>#include <vector>#include <algorithm>using namespace std;const int MAXN = 10005;const int MAXM = 40005;const double EPS = 1e-8;struct Edge {int u, v;Edge(int uu = 0, int vv = 0):u(uu),v(vv){}};int N,M,T,s,t,id[MAXN],idx;vector<int> g[MAXN];vector<Edge> Edges;int father[MAXN];int find(int x) {return x == father[x] ? x : father[x] = find(father[x]);}void unite(int u, int v){    u = find(u), v = find(v);    if(u != v) father[u] = v;}void dfs(int u) {id[u] = idx;for (int i = 0; i < g[u].size(); i++) {int v = g[u][i];if (id[v] != -1) continue;dfs(v);}}double A[1005][1005];double gauss(int n, int s, int t) {for (int i = 0; i < n; i++) {int r;for (r = i; r < n; r++)if (fabs(A[r][i]) >= EPS)break;if (r == n) continue;for (int j = 0; j <= n; j++) swap(A[i][j], A[r][j]);for (int j = n; j > i; j--) A[i][j] /= A[i][i];A[i][i] = 1.0;for (int j = 0; j < n; j++) {if (i == j) continue;if (fabs(A[j][i]) >= EPS) {for (int k = n; k > i; k--)A[j][k] -= A[j][i] * A[i][k];A[j][i] = 0.0;}}}return A[s][n] / A[s][s] - A[t][n] / A[t][t];}void init(int n){    idx = 0;    memset(id, -1, sizeof(id));    memset(A, 0, sizeof(A));    for (int i = 0; i <= n; i++) {        father[i] = i;        g[i].clear();    }    Edges.clear();}int main(void) {    //freopen("input.txt","r",stdin);scanf("%d", &T);while (T--) {scanf("%d %d %d %d", &N, &M, &s, &t);init(N);int u, v, r;while (M--) {scanf("%d %d %d", &u, &v, &r);if (r == 0) {g[u].push_back(v);g[v].push_back(u);}elseEdges.push_back(Edge(u, v));}for (int i = 1; i <= N; i++) {if (id[i] != -1) continue;dfs(i);idx++;}if (id[s] == id[t]) {printf("0.000000\n");continue;}idx++;A[id[s]][idx] = 1;A[id[t]][idx] = -1;A[idx - 1][0] = 1;for (int i = 0, sz = Edges.size(); i < sz; i++) {int u = id[Edges[i].u];int v = id[Edges[i].v];if (u == v) continue;unite(u,v);A[u][u] += 1;A[v][v] += 1;A[u][v] -= 1;A[v][u] -= 1;}if(find(id[s]) != find(id[t]))            puts("inf");        else            printf("%f\n",gauss(idx, id[s], id[t]));}return 0;}