UVA 10828 Back to KernighanRitchie [期望高斯消元]

　　求从1开始，在一个有向图中经过某条边次数的期望。

　　E[p]=sigma(E[t]/deg[t])，t为能到p的点的集合，deg[t]为t的度数。对于起点要额外加1，因为第一次肯定会被运行一次。

　　高斯消元看的Rujia的代码，注意对无解的判断。

#include <string.h>
#include <stdio.h>
#include <vector>
#include <math.h>
#define MAXN 105
using namespace std;
const double eps = 1e-9;
int n, q, deg[MAXN];
int tu, tv;
vector<int> e[MAXN];
double x[MAXN][MAXN];
int inf[MAXN];
void gauss(int n){
    for (int r = 1; r <= n; r++) {
        int maxr = r;
        for (int i = r+1; i <= n; i++)
            if (fabs(x[i][r] > fabs(x[maxr][r]))) maxr = r;
        if (fabs(x[maxr][r]) < eps) continue;
        if (maxr != r)
            for (int i = 1; i <= n+1; i++) swap(x[maxr][i], x[r][i]);
        for (int i = 1; i <= n; i++) if (i != r){
            for (int j = n+1; j >= r; j--)
                x[i][j] -= x[i][r]/x[r][r]*x[r][j];
        }
    }
    memset(inf, 0, sizeof inf);
    for (int i = n; i >= 1; i--) {
        if (fabs(x[i][n+1])>eps && fabs(x[i][i])<eps) inf[i] = 1;
        else if (fabs(x[i][n+1])<eps) x[i][i] = 0;
        else x[i][i] = x[i][n+1]/x[i][i];
        for (int j = i+1; j <= n; j++) {
             if(fabs(x[i][j])>eps && inf[j]) inf[i] = 1;
        }
    }
}
int main(){
    //freopen("test.in","r",stdin);
    int cas = 1;
    while (scanf("%d", &n), n) {
        memset(deg, 0, sizeof deg);
        for (int i = 1; i <= n; i++) e[i].clear();
        while (scanf("%d%d", &tu, &tv), tu||tv) {
            deg[tu]++;
            e[tv].push_back(tu);
        }
        memset(x, 0, sizeof x);
        x[1][n+1] = 1.0;
        for (int i = 1; i <= n; i++) {
            x[i][i] = 1.0;
            for (int j = 0; j < e[i].size(); j++)
                x[i][e[i][j]] -= 1.0/deg[e[i][j]];
        }
        gauss(n);
        printf("Case #%d:\n", cas++);
        scanf("%d", &q);
        while (q--) {
            scanf("%d", &tu);
            if (inf[tu]) printf("infinity\n");
            else printf("%.3f\n", x[tu][tu]);
        }
    }
    return 0;
}