hdu4579 Random Walk

题目：

Yuanfang is walking on a chain. The chain has n nodes numbered from 1 to n. Every second, he can move from node i to node j with probability: 

c(i,j) is an element in a given parameter matrix which is n×m. (1 <= c(i, j) <= 9) 
Yuanfang wants to know the expectation time for him to walk from node 1 to node n.

Input

There are no more than 10 test cases. 
In each case, there are two integers n (2 <= n <= 50000), m (1 <= m <= 5), in the first line, meaning that there are n nodes and the parameter matrix is n×m . There are m integers in each of the next n lines which describe the parameter matrix . 
The input ends with 0 0. 

Output

For each case, output the expectation time for Yuanfang to walk from node 1 to node n in one line. The answer should be rounded to 2 digits after decimal point.

Sample Input

3 11115 21 22 13 22 31 30 0

Sample Output

6.948.75

思路：设dp[i]表示从i到n的步数期望值

那么dp[n]=0

列出n个变元的n个方程，这样就能用高斯消元解出来了。

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3f// 0x3f3f3f3fconst int maxn=5e4+50;const double eps=1e-6;double c[maxn][6],p[maxn][11];double a[maxn][11],b[maxn];double dp[maxn];int main(){    int n,m;    while(~scanf("%d%d",&n,&m)){        if(!n&&!m) break;        for(int i=1;i<=n;i++)            for(int j=1;j<=m;j++)                scanf("%lf",&c[i][j]);        for(int i=1;i<n;i++){            double sum=1,s=0;            for(int j=1;j<=m;j++)                sum+=c[i][j];            for(int j=1;j<=m&&i-j>=1;j++){                p[i][m-j]=0.3*c[i][j]/sum;                s+=p[i][m-j];            }            for(int j=1;j<=m&&i+j<=n;j++){                p[i][m+j]=0.7*c[i][j]/sum;                s+=p[i][m+j];            }            p[i][m]=-s;            b[i]=-1;        }        for(int i=1;i<n;i++){            int L=max(1,i-m);            int R=min(n,i+m);            for(int j=L;j<i;j++){                if(f_abs(p[i][m+j-i])<=eps) continue;                double t=p[i][m+j-i]/a[j][1];                for(int k=1;k<=m+1&&j+k-1<=n;k++){                    p[i][m+j-i+k-1]-=t*a[j][k];                }                b[i]-=t*b[j];            }            for(int j=1;j<=R-i+1;j++)                a[i][j]=p[i][m+j-1];        }        dp[n]=0;        for(int i=n-1;i>=1;i--){            for(int j=2;j<=m+1&&i+j-1<=n;j++)                b[i]-=a[i][j]*dp[i+j-1];            dp[i]=b[i]/a[i][1];        }        printf("%.2f\n",dp[1]);    }    return 0;}