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;
}

UVA 10808 - Rational Resistors(高斯消元+并查集+分数+基尔霍夫定律)