UVa 10828 Back to Kernighan-Ritchie (数学期望 + 高斯消元)

题意：给定一个 n 个结点的有向图，然后从 1 结点出发，从每个结点向每个后继结点的概率是相同的，当走到一个没有后继结点后，那么程序终止，然后问你经过每个结点的期望是次数是多少。

析：假设 i 结点的出度为 di，期望执行次数为 xi，对于一个有 n 个前继结点的 a1, a2, a3 ... an 的结点 i，可以列出方程 xi = xa1/da1 + xa2/da2 + .. + xan/dan，根据每个结点都可以列出一个方程，然后就有 n 个方程，其中结点 1 比较特殊，因为是由它开始的所以看作它有一个前继虚拟结点 0，而 0 只执行一次，所以有 n 个前继结点的 a1, a2, a3 ... an 的结点 1，可以列出方程 x1 = xa1/da1 + xa2/da2 + .. + xan/dan + 1，注意末尾有一个 1，然后就有 n 个方程，然后用高斯消元-约当消元法，注意的是此题可能有无穷多解，和惟一解，多解也就是说最后所以得到的增广矩阵 A[i][i] = 0 && A[i][n] != 0，这样就是无穷大的答案，再就是 A[i][i] = 0 && A[i][n] = 0，这样的话答案就是 0。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 5;
const int maxm = 1e6 + 2;
const LL mod = 1000000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

vector<int> G[maxn];
int d[maxn];
bool ok[maxn];
double A[maxn][maxn];

void solve(){
  for(int i = 0; i < n; ++i){
    int r = i;
    for(int j = i+1; j < n; ++j)
      if(fabs(A[j][i]) > fabs(A[r][i]))  r = j;
    if(fabs(A[r][i]) < eps)  continue;
    if(r != i)  for(int j = i; j <= n; ++j)  swap(A[r][j], A[i][j]);
    
    for(int k = 0; k < n; ++k)  if(k != i)
      for(int j = n; j >= i; --j)  A[k][j] -= A[k][i] / A[i][i] * A[i][j];
  }
}

int main(){
  int kase = 0;
  while(scanf("%d", &n) == 1 && n){
    for(int i = 0; i < n; ++i)  G[i].cl;
    ms(d, 0);  ms(ok, 0);  ms(A, 0);  A[0][n] = 1.;
    int a, b;
    while(scanf("%d %d", &a, &b) == 2 && a+b){
      G[b-1].pb(a-1);  ++d[a-1];
    }
    for(int i = 0; i < n; ++i){
      A[i][i] = 1.;
      for(int j = 0; j < G[i].sz; ++j)
        A[i][G[i][j]] -= 1. / d[G[i][j]];
    }
    solve();
    for(int i = n-1; i >= 0; --i){
      if(fabs(A[i][i]) < eps && fabs(A[i][n]) > eps){  ok[i] = true;  continue; }
      for(int j = i+1; j < n; ++j)
        if(fabs(A[i][j]) > eps && ok[j])  ok[i] = true;
    }
    printf("Case #%d:
", ++kase);
    scanf("%d", &m);
    int x;
    while(m-- && scanf("%d", &x) == 1)
      if(ok[x-1])  puts("infinity");
      else if(fabs(A[x-1][x-1]) < eps)  printf("0.000
");
      else printf("%.3f
", A[x-1][n] / A[x-1][x-1]);
  }
  return 0;
}