UVa 10828 back to Kernighan-ritchie Gaussian elimination + probability DP

Topic Source: UVa 10828 back to Kernighan-ritchie

Test instructions: Starting from 1 each time the equal probability from one point to the point of his neighbor has to go to not go stop seeking to stop the expectation of each point

Idea: To write the equation elimination equation has a unique solution to many solutions without solution of the ring has been in the loop can not stop the calculation infinity from 1 can not reach the point expected to be 0

#include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <cmath&
Gt
using namespace Std;
const int MAXN = 110;
Const double EPS = 1e-8;
typedef double MATRIX[MAXN][MAXN];
Vector <int> G[MAXN];
int D[MAXN];
int INF[MAXN];
	void Gauss_jordan (Matrix A, int n) {int I, J, R, K;
		for (i = 0; i < n; i++) {r = i;
		for (j = i+1; J < N; j + +) if (Fabs (A[j][i]) > Fabs (A[r][i])) R = J;
			if (Fabs (a[r][i) > EPS) {if (r! = i) for (j = 0; J <= N; j + +) Swap (A[r][j], a[i][j]); for (k = 0; k < n; k++) {if (k! = i) {for (j = n; j >= i; j--) {a[k][j]-= a[k][i]/a[i][
					I] * a[i][j];
}}}}}} Matrix A;
	int main () {int cas = 1;
	int n;
		while (scanf ("%d", &n) && N) {memset (d, 0, sizeof (d));
		for (int i = 0; i < n; i++) g[i].clear ();
		int u, v; while (scanf ("%d%d", &u, &v) && (u| |
			V)) {u--;
			v--; G[v].push_bacK (U);
		d[u]++;
		} memset (A, 0, sizeof (a));
			for (int i = 0; i < n; i++) {a[i][i] = 1;
			for (int j = 0; J < G[i].size (); j + +) A[i][g[i][j]]-= 1.0/d[g[i][j]];
		if (i = = 0) A[i][n] = 1;
		
		} Gauss_jordan (A, N);
		memset (INF, 0, sizeof (INF));
			for (int i = n-1; I >= 0; i--) {if (Fabs (A[i][i]) < EPS && fabs (a[i][n]) > EPS) inf[i] = 1;
		for (int j = i+1; J < N; j + +) if (Fabs (A[i][j]) > EPS && inf[j]) inf[i] = 1;
		} int q;
		scanf ("%d", &q);
		printf ("Case #%d:\n", cas++);
			while (q--) {int x;
			scanf ("%d", &x);
			x--;
				if (Inf[x]) {puts ("infinity");
			Continue
				} if (Fabs (A[x][x]) < EPS) {puts ("0.000");
			Continue
		} printf ("%.3lf\n", A[x][n]/a[x][x]);
}} return 0; }