UVA-10828 (概率期望+高斯消元)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
#define ll long long;
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
    for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + ‘0‘);
    putchar(‘\n‘);
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+2000;
const int maxn=110;
const double eps=1e-8;

int n,u,v,num[110];
vector<int>ve[110];

double a[maxn][maxn],ans[maxn];
int in[maxn],d[maxn];

inline void Init()
{
    For(i,1,n)num[i]=ve[i].size();
    For(i,1,n+1)For(j,1,n+1)a[i][j]=0;
    for(int i=1;i<=n;i++)
    {
        a[i][i]=1;
        for(int j=0;j<num[i];j++)
        {
            a[i][ve[i][j]]-=1.0/d[ve[i][j]];
        }
        if(i==1)a[i][n+1]=1;
    }
}
void gauss()
{
    for(int i=1;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=0;j<=n+1;j++)swap(a[r][j],a[i][j]);
        for(int k=1;k<=n;k++)
        {
            if(k==i)continue;
            for(int j=n+1;j>=i;j--)a[k][j]-=a[k][i]/a[i][i]*a[i][j];
        }
     }
}
int main()
{
    int Case=0;
    while(1)
    {

        read(n);
        if(!n)break;
        printf("Case #%d:\n",++Case);
        For(i,1,n)ve[i].clear(),d[i]=0;
        while(1)
        {
            read(u);read(v);
            if(!u&&!v)break;
            d[u]++;
            ve[v].push_back(u);
        }
        Init();
        gauss();
        mst(in,0);
        for(int i=n;i>0;i--)
        {
            if(fabs(a[i][i])<eps&&fabs(a[i][n+1])>eps)in[i]=1;
            for(int j=i+1;j<=n;j++)
                if(fabs(a[i][j])>eps&&in[j])in[i]=1;
        }
        int q,x;
        read(q);
        while(q--)
        {
            read(x);
            if(in[x])printf("infinity\n");
            else printf("%.3lf\n",fabs(a[x][x])<eps? 0.0:a[x][n+1]/a[x][x]);
        }
    }
    return 0;
}