HDU 4579 Random Walk （解方程组）

Random Walk

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65535/65536 K (Java/Others)
Total Submission(s): 81    Accepted Submission(s): 35

Problem Description

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 1 1 1 1 5 2 1 2 2 1 3 2 2 3 1 3 0 0

 

 

Sample Output

6.94 8.75

 

 

Source

2013ACM-ICPC杭州赛区全国邀请赛

 

 

 

 

 

这题就是列出方程。

 

然后高斯消元解方程。

 

如果是一般的方程，高斯消元需要O(n*n)

 

但是这题的矩阵很特别

 

 

很快就可以消掉了。

用dp[i]表示i到n的期望。

dp[n]=0;

 

 

对于第i个方程，最多只有2*m+1个系数是不为0的。

 

 

 

 

 

 1 /* **********************************************

 2 Author      : kuangbin

 3 Created Time: 2013/8/12 20:28:58

 4 File Name   : F:\2013ACM练习\比赛练习\2013杭州邀请赛重现\1004.cpp

 5 *********************************************** */

 6 

 7 #include <stdio.h>

 8 #include <string.h>

 9 #include <iostream>

10 #include <algorithm>

11 #include <vector>

12 #include <queue>

13 #include <set>

14 #include <map>

15 #include <string>

16 #include <math.h>

17 #include <stdlib.h>

18 using namespace std;

19 const int MAXN = 50010;

20 double c[MAXN][10];

21 double p[MAXN][20];

22 double a[MAXN][10];

23 double b[MAXN];

24 double dp[MAXN];

25 

26 int main()

27 {

28     //freopen("in.txt","r",stdin);

29     //freopen("out.txt","w",stdout);

30     int n,m;

31     while( scanf("%d%d",&n,&m) == 2 )

32     {

33         if(n == 0 && m == 0)break;

34         for(int i = 1;i <= n;i++)

35             for(int j = 1;j <= m;j++)

36                 scanf("%lf",&c[i][j]);

37         for(int i = 1;i < n;i++)

38         {

39             double sum = 0;

40             for(int j = 1;j <= m;j++)

41                 sum += c[i][j];

42             double s = 0;

43             for(int j = 1;j <= m && i-j >= 1;j++)

44             {

45                 p[i][m-j] = 0.3*c[i][j]/(1+sum);

46                 s += p[i][m-j];

47             }

48             for(int j = 1;j <= m && i+j <= n;j++)

49             {

50                 p[i][m+j] = 0.7*c[i][j]/(1+sum);

51                 s += p[i][m+j];

52             }

53             p[i][m] = -s;

54             b[i] = -1;

55         }

56         for(int i = 1;i <= m+1 && i <= n;i++)

57             a[1][i] = p[1][m+i-1];

58         for(int i = 2;i < n;i++)

59         {

60             int end = min(i+m,n);

61             int start = max(1,i-m);

62             

63             for(int j = start;j < i;j++)

64                 if(fabs(p[i][m+j-i]) > 1e-6)

65                 {

66                     double t = p[i][m+j-i]/a[j][1];

67                     for(int k = 1; k <= m+1 && j+k-1 <= n ;k++)

68                     {

69                         p[i][m+j-i+k-1] -= t*a[j][k];

70                     }

71                     b[i] -= t*b[j];

72                 }

73             for(int j = 1;j <= end-i+1;j++)

74                 a[i][j] = p[i][m+j-1];

75 

76         }

77         dp[n] = 0;

78         for(int i = n-1;i >= 1;i--)

79         {

80             for(int j = 2;j <= m+1 && i+j-1 <= n;j++)

81                 b[i] -= dp[i+j-1] * a[i][j];

82             dp[i] = b[i]/a[i][1];

83         }

84         printf("%.2f\n",dp[1]);

85     }

86     return 0;

87 }