标签:div cas print continue nta namespace eps ons media
‘Snakes and Ladders‘ or ‘Shap-Ludu‘ is a game commonly played in Bangladesh. The game is so common that it would be tough to find a person who hasn‘t played it. But those who haven‘t played it (unlucky of course!) the rules are as follows.
Now given a board, you have to find the expected number of times you need to throw the dice to win the game. The cases will be given such that a result will be found.
Input
Input starts with an integer T (≤ 105), denoting the number of test cases.
The first line of a case is a blank line. The next line gives you an integer n denoting the number of snakes and ladders. Each of the next n lines contain two integers a and b (1 ≤ a, b ≤ 100, a ≠ b). If a < b, it means that there is a ladder which takes you from a to b. If a > b, it means that there is a snake which takes you from a to b. Assume that the given board follows the above restrictions.
Output
For each case of input, print the case number and the expected number of times you need to throw the dice. Errors less than 10-6 will be ignored.
Sample Input
2
14
4 42
9 30
16 8
14 77
32 12
37 58
47 26
48 73
62 19
70 89
71 67
80 98
87 24
96 76
0
Sample Output
Case 1: 31.54880806
Case 2: 33.0476190476
有待理解:
1 #include<bits/stdc++.h> 2 #define EPS 1e-8 3 using namespace std; 4 5 int x,kase; 6 int a[110]; 7 double dp[110][110]; 8 9 void Gauss(int n,int m){ 10 int col,row,mxr; 11 for(col=row=1;row<=n&&col<=m;row++,col++){ 12 mxr=row; 13 for(int i=row+1;i<=n;i++) if(fabs(dp[i][col])>fabs(dp[mxr][col])) mxr=i; 14 if(mxr!=row) swap(a[row],a[mxr]); 15 if(fabs(dp[row][col])<EPS){ row--; continue; } 16 for(int i=1;i<=n;i++){ 17 if(i!=row&&fabs(dp[i][col])>EPS){ 18 for(int j=m;j>=col;j--) dp[i][j]-=dp[row][j]/dp[row][col]*dp[i][col]; 19 } 20 } 21 } 22 row--; 23 for(int i=row;i>=1;i--){ 24 for(int j=i+1;j<=row;j++) dp[i][m]-=dp[j][m]*dp[i][j]; 25 dp[i][m]/=dp[i][i]; 26 } 27 } 28 29 void Solve(){ 30 for(int i=1;i<100;i++){ 31 if(a[i]){ 32 dp[i][a[i]]=-1; 33 dp[i][101]=0; 34 dp[i][i]=1; 35 } 36 else{ 37 int cnt=0; 38 for(int j=1;i+j<=100&&j<=6;j++){ 39 cnt++; 40 dp[i][i+j]=-1; 41 } 42 dp[i][i]=cnt; 43 dp[i][101]=6; 44 } 45 } 46 dp[100][100]=1; 47 dp[100][101]=0; 48 Gauss(100,101); 49 } 50 51 int main() 52 { cin>>kase; 53 for(int t=1;t<=kase;t++){ 54 cin>>x; 55 memset(a,0,sizeof(a)); 56 memset(dp,0,sizeof(dp)); 57 for(int i=1;i<=x;i++){ 58 int u,v; 59 cin>>u>>v; 60 a[u]=v; 61 } 62 Solve(); 63 printf("Case %d: %lf\n",t,dp[1][101]); 64 } 65 return 0; 66 }
Snakes and Ladders LightOJ - 1151
标签:div cas print continue nta namespace eps ons media
原文地址:http://www.cnblogs.com/zgglj-com/p/7806934.html