lightoj 1151 Snakes and Ladders 期望 高斯消元

设出每个点的期望r[i]。

如果i点会瞬移到to[i]，则r[i] = r[to[i]]

否则，r[i] = (Σr[j] + (6-cnt)*r[i])/6 + 1 (j为可以往右走的位置，cnt表示其数目，至多6个)

值得注意的是，这个位置最后可能会发现出界了然后原地呆一轮，这种情况一样有贡献，需要注意。

然后列出方程，高斯消元即可。

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 105;
const double eps = 1e-9;
double a[MAXN][MAXN]; 
int T,cas,n,equ,var;
int to[MAXN];

int gauss(int n,int m)
{
    int col,row,mxr;
    for(row=col=1;row<=n&&col<=m;row++,col++)
    {
        mxr=row;
        for(int i=row+1;i<=n;i++)
            if(fabs(a[i][col])>fabs(a[row][col]))
                mxr=i;
        if(mxr!=row) swap(a[row],a[mxr]);
        if(fabs(a[row][col])<eps)
        {
            row--;
            continue;
        }
        for(int i=1;i<=n;i++) //消成上三角矩阵
            if(i!=row&&fabs(a[i][col])>eps)
                for(int j=m;j>=col;j--)
                    a[i][j]-=a[row][j]/a[row][col]*a[i][col];
    }
    row--;
    for(int i=row;i>=1;i--) //回代成对角矩阵
    {
        for(int j=i+1;j<=row;j++)
            a[i][m]-=a[j][m]*a[i][j];
        a[i][m]/=a[i][i];
    }
    return row;
}
int main()
{
    for (scanf("%d",&T);T != 0;T--)
    {
        cas++;
        for (int i = 1;i <= 100;i++)
        {
            for (int j = 1;j <= 101;j++)
                a[i][j] = 0;
            to[i] = 0;
        }
        scanf("%d",&n);    
        int ta,tb;
        for (int i = 1;i <= n;i++)
        {
            scanf("%d%d",&ta,&tb);
            to[ta] = tb;
        }
        for (int i = 1;i <= 99;i++)
        {
            if (to[i] != 0)
            {
                a[i][to[i]] = -1.0;
                a[i][101] = 0.0;
                a[i][i] = 1.0;
            }    
            else
            {
                int cnt = min(6,100 - i);
                a[i][i] = 1.0 - (6.0 - cnt) / 6.0;
                for (int j = i + 1;j <= min(100,i + 6);j++)
                    a[i][j] = -1.0 / 6;
                a[i][101] = 1.0; 
            }
        }
        a[100][100] = 1.0;
        a[100][101] = 0.0;
        gauss(100,101);
        printf("Case %d: %.6lf\n",cas,a[1][101]);
    }
    return 0;
}