【HDOJ】4326 Game

1. 题目描述
一个长度为n个队列，每次取队头的4个人玩儿游戏，每个人等概率赢得比赛。胜者任然处在队头，然而败者按照原顺序依次排在队尾。连续赢得m场比赛的玩家赢得最终胜利。
求第k个人赢得最终胜利的概率。

2. 基本思路
显然是个概率DP，dp[i][j]表示第1个玩家已经连续赢得i局比赛时，第j个人赢得最终胜利的概率。所求极为dp[0][k]。
\[
dp[m][j] = \begin{cases}
\begin{aligned}
&1,  j=1 \\
&0,  j>1
\end{aligned}
\end{cases}
\]
$dp[i][j] =$
\[
\quad \left\{ \begin{aligned}
    &\frac{1}{4}dp[i+1][j] + \frac{3}{4}dp[1][n-2], &j=1 \\
    &\frac{1}{4}dp[i+1][n-2] + \frac{1}{4}dp[1][1] + \frac{2}{4}dp[1][n-1], &j=2 \\
    &\frac{1}{4}dp[i+1][n-3] + \frac{1}{4}dp[1][n-1] + \frac{1}{4}dp[1][1] + \frac{1}{4}dp[1][n], &j=3 \\
    &\frac{1}{4}dp[i+1][n] + \frac{2}{4}dp[1][n] + \frac{1}{4}dp[1][1], &j=4 \\
    &\frac{3}{4}dp[1][j-3] + \frac{1}{4}dp[i+1][j-3], &j>4
\end{aligned}
\right .
\]
因为找不到一个有效的常量，因此考虑解n*m元方程组。方法是高斯消元。

3. 代码

 /* 4326 */
 #include <iostream>
 #include <sstream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <bitset>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 const int maxn = ;
 const double eps = 1e-;
 typedef double mat[maxn][maxn];
 double x[maxn];
 mat g;
 int n, m, k;

 void gauss_elimination(mat& g, int n) {
     int r;

     rep(i, , n) {
         r = i;
         rep(j, i+, n) {
             if (fabs(g[j][i]) > fabs(g[r][i]))
                 r = j;
         }
         if (r != i) {
             rep(j, , n+)
                 swap(g[r][j], g[i][j]);
         }

         rep(k, i+, n) {
             if (fabs(g[i][i]) < eps)
                 continue;
             double f = g[k][i] / g[i][i];
             rep(j, i, n+)
                 g[k][j] -= f * g[i][j];
         }
     }

     per(i, , n) {
         rep(j, i+, n)
             g[i][n] -= g[j][n] * g[i][j];
         g[i][n] /= g[i][i];
     }
 }

 void add(int ridx, int i, int j, double val) {
     if (i == m) {
         if (j == )
             g[ridx][i*n+j-] -= val;
         return ;
     }

     g[ridx][i*n+j-] += val;
 }

 void solve() {
     int idx = ;

     memset(g, , sizeof(g));

     rep(i, , m) {
         rep(j, , n+) {
             add(idx, i, j, 1.0);
             if (j == ) {
                 add(idx, i+, j, -0.25);
                 add(idx, , n-, -0.75);
             } else if (j == ) {
                 add(idx, i+, n-, -0.25);
                 add(idx, , , -0.25);
                 add(idx, , n-, -0.5);
             } else if (j == ) {
                 add(idx, i+, n-, -0.25);
                 add(idx, , n-, -0.25);
                 add(idx, , , -0.25);
                 add(idx, , n, -0.25);
             } else if (j == ) {
                 add(idx, i+, n, -0.25);
                 add(idx, , n, -0.5);
                 add(idx, , , -0.25);
             } else {
                 add(idx, , j-, -0.75);
                 add(idx, i+, j-, -0.25);
             }
             ++idx;
         }
     }

     gauss_elimination(g, idx);
     double ans = g[k-][idx];
     printf("%.6lf\n", ans);
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     int t;

     scanf("%d", &t);
     rep(tt, , t+) {
         scanf("%d%d%d", &n, &m, &k);
         printf("Case #%d: ", tt);
         solve();
     }

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }