UVA 10828，uva10828

UVA 10828，uva10828
UVA 10828 - Back to Kernighan-Ritchie

題目連結

題意：給圖一個流程圖，有結點的流程，每次進入下一個流程機率是均等的，有q次詢問，求出每次詢問結點的執行期望

思路：高斯消元，每個結點的期望等於所有前趨結點的期望/出度的和，由於存在無限迴圈的情況，不能直接遞推，利用高斯消元去做，判斷無解的情況既為無限迴圈，注意如果一個式自xi為0，但是xn也為0，xi值應該是0，表示無法到達

代碼：

#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;}