题目链接：uva 10808 - Rational Resistors
题目大意：给出一个博阿含n个节点，m条导线的电阻网络，求节点a和b之间的等效电阻。
解题思路：基尔霍夫定律，任何一点的电流向量为0。就是说有多少电流流入该节点，就有多少电流流出。
对于每次询问的两点间等效电阻，先判断说两点是否联通，不连通的话绝逼是1/0（无穷大）。联通的话，将同一个联通分量上的节点都扣出来，假设电势作为变元，然后根据基尔霍夫定律列出方程，因为对于每个节点的电流向量为0，所以每个节点都有一个方程，所有与该节点直接连接的都会有电流流入，并且最后总和为0，（除了a，b两点，一个为1，一个为-1）。用高斯消元处理，但是这样列出的方程组不能准确求出节点的电势，只能求出各个节点之间电势的关系。所以我们将a点的电势置为0，那么用求出的b点电势减去0就是两点间的电压，又因为电流设为1，所以等效电阻就是电压除以电流。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long type;
struct Fraction {
type member; // 分子；
type denominator; // 分母；
Fraction (type member = 0, type denominator = 1);
void operator = (type x) { this->set(x, 1); }
Fraction operator * (const Fraction& u);
Fraction operator / (const Fraction& u);
Fraction operator + (const Fraction& u);
Fraction operator - (const Fraction& u);
Fraction operator *= (const Fraction& u) { return *this = *this * u; }
Fraction operator /= (const Fraction& u) { return *this = *this / u; }
Fraction operator += (const Fraction& u) { return *this = *this + u; }
Fraction operator -= (const Fraction& u) { return *this = *this - u; }
void set(type member, type denominator);
};
inline type gcd (type a, type b) {
return b == 0 ? (a > 0 ? a : -a) : gcd(b, a % b);
}
inline type lcm (type a, type b) {
return a / gcd(a, b) * b;
}
/*Code*/
/////////////////////////////////////////////////////
const int maxn = 105;
typedef long long ll;
typedef Fraction Mat[maxn][maxn];
int N, M, f[maxn];
Mat G, A;
bool cmp (Fraction& a, Fraction& b) {
return a.member * b.denominator < b.member * a.denominator;
}
inline int getfar (int x) {
return x == f[x] ? x : f[x] = getfar(f[x]);
}
inline void link (int u, int v) {
int p = getfar(u);
int q = getfar(v);
f[p] = q;
}
void init () {
scanf("%d%d", &N, &M);
for (int i = 0; i < N; i++) {
f[i] = i;
for (int j = 0; j < N; j++)
G[i][j] = 0;
}
int u, v;
ll R;
for (int i = 0; i < M; i++) {
scanf("%d%d%lld", &u, &v, &R);
if (u == v)
continue;
link(u, v);
G[u][v] += Fraction(1, R);
G[v][u] += Fraction(1, R);
}
}
Fraction gauss_elimin (int u, int v, int n) {
/*
printf("\n");
for (int i = 0; i < n; i++) {
for (int j = 0; j <= n; j++)
printf("%lld/%lld ", A[i][j].member, A[i][j].denominator);
printf("\n");
}
*/
for (int i = 0; i < n; i++) {
int r;
for (int j = i; j < n; j++)
if (A[j][i].member) {
r = j;
break;
}
if (r != i) {
for (int j = 0; j <= n; j++)
swap(A[i][j], A[r][j]);
}
if (A[i][i].member == 0)
continue;
for (int j = i + 1; j < n; j++) {
Fraction t = A[j][i] / A[i][i];
for (int k = 0; k <= n; k++)
A[j][k] -= A[i][k] * t;
}
}
for (int i = n-1; i >= 0; i--) {
for (int j = i+1; j < n; j++) {
if (A[j][j].member)
A[i][n] -= A[i][j] * A[j][n] / A[j][j];
}
}
/*
Fraction U = A[u][n] / A[u][u];
printf("%lld/%lld!\n", A[u][n].member, A[u][n].denominator);
printf("%lld/%lld!\n", A[u][u].member, A[u][u].denominator);
printf("%lld/%lld\n", U.member, U.denominator);
Fraction V = A[v][n] / A[v][v];
printf("%lld/%lld\n", V.member, V.denominator);
*/
return A[u][n] / A[u][u] - A[v][n] / A[v][v];
}
Fraction solve (int u, int v) {
int n = 0, hash[maxn];
int hu, hv;
for (int i = 0; i < N; i++) {
if (i == u)
hu = u;
if (i == v)
hv = v;
if (getfar(i) == getfar(u))
hash[n++] = i;
}
n++;
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= n; j++)
A[i][j] = 0;
}
for (int i = 0; i < n - 1; i++) {
for (int j = 0; j < n - 1; j++) {
if (i == j)
continue;
int p = hash[i];
int q = hash[j];
A[i][i] += G[p][q];
A[i][j] -= G[p][q];
}
}
A[hu][n] = 1;
A[hv][n] = -1;
A[n-1][0] = 1;
return gauss_elimin (hu, hv, n);
}
int main () {
int cas;
scanf("%d", &cas);
for (int kcas = 1; kcas <= cas; kcas++) {
init();
int Q, u, v;
scanf("%d", &Q);
printf("Case #%d:\n", kcas);
for (int i = 0; i < Q; i++) {
scanf("%d%d", &u, &v);
printf("Resistance between %d and %d is ", u, v);
if (getfar(u) == getfar(v)) {
Fraction ans = solve(u, v);
printf("%lld/%lld\n", ans.member, ans.denominator);
} else
printf("1/0\n");
}
printf("\n");
}
return 0;
}
/////////////////////////////////////////////////////
Fraction::Fraction (type member, type denominator) {
this->set(member, denominator);
}
Fraction Fraction::operator * (const Fraction& u) {
type tmp_p = gcd(member, u.denominator);
type tmp_q = gcd(u.member, denominator);
return Fraction( (member / tmp_p) * (u.member / tmp_q), (denominator / tmp_q) * (u.denominator / tmp_p) );
}
Fraction Fraction::operator / (const Fraction& u) {
type tmp_p = gcd(member, u.member);
type tmp_q = gcd(denominator, u.denominator);
return Fraction( (member / tmp_p) * (u.denominator / tmp_q), (denominator / tmp_q) * (u.member / tmp_p));
}
Fraction Fraction::operator + (const Fraction& u) {
type tmp_l = lcm (denominator, u.denominator);
return Fraction(tmp_l / denominator * member + tmp_l / u.denominator * u.member, tmp_l);
}
Fraction Fraction::operator - (const Fraction& u) {
type tmp_l = lcm (denominator, u.denominator);
return Fraction(tmp_l / denominator * member - tmp_l / u.denominator * u.member, tmp_l);
}
void Fraction::set (type member, type denominator) {
if (denominator == 0) {
denominator = 1;
member = 0;
}
if (denominator < 0) {
denominator = -denominator;
member = -member;
}
type tmp_d = gcd(member, denominator);
this->member = member / tmp_d;
this->denominator = denominator / tmp_d;
}uva 10808 - Rational Resistors(基尔霍夫定律+高斯消元),布布扣,bubuko.com
时间： 2024-12-01 07:39:10