【期望DP+高斯消元】 UVA 10828 Back to Kernighan-Ritchie

通道

题意：从1开始 每次等概率从一个点到和他相邻的点 有向 走到不能走停止 求停止时每个点的期望

思路：白书P156

代码：

#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;  
}