10828-back to kernighan-Ritchie (probability + Gaussian elimination)

Ultraviolet A 10828-back to kernighan-Ritchie

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A flowchart with nodes is shown. The probability of entering the next process is equal. Q queries are performed to find the execution expectation of each query node.

Idea: Gaussian deyuan. The expectation of each node is equal to the sum of expectation/exit of all the frontend nodes. Due to the infinite loop, it is not possible to directly recursion, use Gaussian deyuan to determine that the unsolvable condition is an infinite loop. Note that if a formula from Xi is 0, but XN is also 0, and the XI value is 0, it indicates that it cannot be reached.

Code:

#include <cstdio>#include <cstring>#include <cmath>#include <vector>using namespace std;const int N = 105;const double eps = 1e-9;int n, d[N], inf[N];double a[N][N];vector<int> pre[N];void build() {    int u, v;    memset(d, 0, sizeof(d));    for (int i = 0; i < n; i++)pre[i].clear();    while (~scanf("%d%d", &u, &v) && u) {u--; v--; d[u]++;pre[v].push_back(u);    }    memset(a, 0, sizeof(a));    for (int i = 0; i < n; i++) {a[i][i] = 1;for (int j = 0; j < pre[i].size(); j++)    a[i][pre[i][j]] = -1.0 / d[pre[i][j]];if (i == 0) a[i][n] = 1;    }}void gauss() {    for (int i = 0; i < n; i++) {int k = i;for (;k < n; k++)    if (fabs(a[k][i]) > eps) break;if (k == n) continue;for (int j = 0; j <= n; j++) swap(a[k][j], a[i][j]);for (int j = 0; j < n; j++) {    if (i == j) continue;    if (fabs(a[k][i]) > eps) {double x = a[j][i] / a[i][i];for (int k = i; k <= n; k++)    a[j][k] -= x * a[i][k];    }}    }}void get_inf() {    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 main() {    int cas = 0;    while (~scanf("%d", &n) && n) {build();gauss();get_inf();int q, node;scanf("%d", &q);printf("Case #%d:\n", ++cas);while (q--) {    scanf("%d", &node);    node--;    if (inf[node]) printf("infinity\n");    else printf("%.3lf\n", fabs(a[node][node]) < eps ? 0 : a[node][n] / a[node][node]);}    }    return 0;}