首页 > 代码库 > UVA 10808 - Rational Resistors(高斯消元+并查集+分数+基尔霍夫定律)
UVA 10808 - Rational Resistors(高斯消元+并查集+分数+基尔霍夫定律)
UVA 10808 - Rational Resistors
题意:给定一些结点,有一些电阻,电阻分布在边上,给定一个电路图,每次询问两点,求这两点间的等效电阻
思路:根据基尔霍夫定律,任意一点的电流向量为0,这样就能设每个结点的电势,列出方程,利用高斯消元求解,对于无解的情况,肯定是两点不能连通,这个可以利用并查集判断。
此外这题有个很坑的地方啊,就是高斯消元的姿势不够优美就会爆long long
代码:
#include <cstdio> #include <cstring> #include <algorithm> #include <vector> using namespace std; typedef long long ll; struct Frac { ll a, b; Frac() {a = 0; b = 1;} Frac(ll a, ll b) {this->a = a; this->b = b; deal();} void init() {a = 0; b = 1;} ll gcd(ll a, ll b) { while (b) { ll tmp = a % b; a = b; b = tmp; } return a; } void deal() { ll d = gcd(a, b); a /= d; b /= d; if (b < 0) { a = -a; b = -b; } } Frac operator + (Frac c) { Frac ans; ans.a = a * c.b + b * c.a; ans.b = b * c.b; ans.deal(); return ans; } Frac operator - (Frac c) { Frac ans; ans.a = a * c.b - b * c.a; ans.b = b * c.b; ans.deal(); return ans; } Frac operator * (Frac c) { Frac ans; ans.a = a * c.a; ans.b = b * c.b; ans.deal(); return ans; } Frac operator / (Frac c) { Frac ans; ans.a = a * c.b; ans.b = b * c.a; ans.deal(); return ans; } void operator += (Frac c) {*this = *this + c;} void operator -= (Frac c) {*this = *this - c;} void operator *= (Frac c) {*this = *this * c;} void operator /= (Frac c) {*this = *this / c;} bool operator > (Frac c) {return a * c.b > b * c.a;} bool operator == (Frac c) { return a * c.b == b * c.a;} bool operator < (Frac c) {return !(*this < c && *this == c);} bool operator >= (Frac c) {return !(*this < c);} bool operator <= (Frac c) {return !(*this > c);} bool operator != (Frac c) {return !(*this == c);} bool operator != (ll c) {return *this != Frac(c, 1);} void operator = (ll c) {this->a = c; this->b = 1;} }; const int N = 20; int t, n, m, parent[N], node[N]; Frac g[N][N], A[N][N]; int find(int x) { return x == parent[x] ? x : parent[x] = find(parent[x]); } Frac gauss(int n, int u, int v) { for (int i = 0; i < n; i++) { int r; for (r = i; r < n; r++) if (A[r][i] != 0) 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 (A[j][i] != 0) { for (int k = n; k > i; k--) A[j][k] -= A[j][i] * A[i][k]; A[j][i] = 0; } } } return A[u][n] / A[u][u] - A[v][n] / A[v][v]; } Frac solve(int u, int v) { int tu, tv, tn = 0; for (int i = 0; i < n; i++) { if (i == u) tu = tn; if (i == v) tv = tn; if (find(u) == find(i)) node[tn++] = i; } tn++; for (int i = 0; i < tn; i++) for (int j = 0; j <= tn; j++) A[i][j].init(); for (int i = 0; i < tn - 1; i++) { for (int j = 0; j < tn - 1; j++) { if (i == j) continue; int u = node[i], v = node[j]; A[i][i] += g[u][v]; A[i][j] -= g[u][v]; } } A[tu][tn] = 1; A[tv][tn] = -1; A[tn - 1][0] = 1; return gauss(tn, tu, tv); } int main() { int cas = 0; scanf("%d", &t); while (t--) { scanf("%d%d", &n, &m); for (int i = 0; i < n; i++) { parent[i] = i; for (int j = 0; j < n; j++) g[i][j].init(); } int u, v; ll r; while (m--) { scanf("%d%d%lld", &u, &v, &r); if (u == v) continue; g[u][v] += Frac(1, r); g[v][u] += Frac(1, r); int pu = find(u); int pv = find(v); if (pu != pv) parent[pu] = pv; } scanf("%d", &m); printf("Case #%d:\n", ++cas); while (m--) { scanf("%d%d", &u, &v); int pu = find(u); int pv = find(v); printf("Resistance between %d and %d is ", u, v); if (pu != pv) printf("1/0\n"); else { Frac ans = solve(u, v); printf("%lld/%lld\n", ans.a, ans.b); } } printf("\n"); } return 0; }
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