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HDU 5006 Resistance
思路：这题由于数据是随机的。。电阻不是1就是0，就可以先缩点，把电阻为0的那些边缩掉，只考虑有电阻的边，这样的话缩下来点数就不多了，就可以利用高斯消元+基尔霍夫定律去搞了
代码：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;const int N = 10005;
const int M = 40005;
const double eps = 1e-8;int t, en;struct Edge {int u, v;Edge(int u = 0, int v = 0) {this->u = u;this->v = v;}
} E[M];int n, m, s, e, id[N], idx;
vector<int> g[N];
int parent[N];int find(int x) {return x == parent[x] ? x : parent[x] = find(parent[x]);
}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];void gauss(int n, int s, int e) {if (find(s) != find(e)) {printf("inf\n");return;}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;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;}}}printf("%.6lf\n", A[s][n] / A[s][s] - A[e][n] / A[e][e]);
}int main() {scanf("%d", &t);while (t--) {memset(id, -1, sizeof(id));en = 0;scanf("%d%d%d%d", &n, &m, &s, &e);int u, v, r;for (int i = 0; i <= n; i++) {parent[i] = i;g[i].clear();}while (m--) {scanf("%d%d%d", &u, &v, &r);if (r == 0) {g[u].push_back(v);g[v].push_back(u);} elseE[en++] = Edge(u, v);}idx = 0;for (int i = 1; i <= n; i++) {if (id[i] != -1) continue;dfs(i);idx++;}if (id[s] == id[e]) {printf("0.000000\n");continue;}memset(A, 0, sizeof(A));idx++;A[id[s]][idx] = 1;A[id[e]][idx] = -1;A[idx - 1][0] = 1;for (int i = 0; i < en; i++) {int u = id[E[i].u];int v = id[E[i].v];if (u == v) continue;int pu = find(u);int pv = find(v);if (pu != pv) parent[pu] = pv;A[u][u] += 1;A[v][v] += 1;A[u][v] -= 1;A[v][u] -= 1;}gauss(idx, id[s], id[e]);}return 0;
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