HDOJ4326 Game

Posted 2020-06-18 Bombe

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

  1 /* 4326 */
  2 #include <iostream>
  3 #include <sstream>
  4 #include <string>
  5 #include <map>
  6 #include <queue>
  7 #include <set>
  8 #include <stack>
  9 #include <vector>
 10 #include <deque>
 11 #include <bitset>
 12 #include <algorithm>
 13 #include <cstdio>
 14 #include <cmath>
 15 #include <ctime>
 16 #include <cstring>
 17 #include <climits>
 18 #include <cctype>
 19 #include <cassert>
 20 #include <functional>
 21 #include <iterator>
 22 #include <iomanip>
 23 using namespace std;
 24 //#pragma comment(linker,"/STACK:102400000,1024000")
 25 
 26 #define sti                set<int>
 27 #define stpii            set<pair<int, int> >
 28 #define mpii            map<int,int>
 29 #define vi                vector<int>
 30 #define pii                pair<int,int>
 31 #define vpii            vector<pair<int,int> >
 32 #define rep(i, a, n)     for (int i=a;i<n;++i)
 33 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 34 #define clr                clear
 35 #define pb                 push_back
 36 #define mp                 make_pair
 37 #define fir                first
 38 #define sec                second
 39 #define all(x)             (x).begin(),(x).end()
 40 #define SZ(x)             ((int)(x).size())
 41 #define lson            l, mid, rt<<1
 42 #define rson            mid+1, r, rt<<1|1
 43 
 44 const int maxn = 105;
 45 const double eps = 1e-8;
 46 typedef double mat[maxn][maxn];
 47 double x[maxn];
 48 mat g;
 49 int n, m, k;
 50 
 51 void gauss_elimination(mat& g, int n) {
 52     int r;
 53 
 54     rep(i, 0, n) {
 55         r = i;
 56         rep(j, i+1, n) {
 57             if (fabs(g[j][i]) > fabs(g[r][i]))
 58                 r = j;
 59         }
 60         if (r != i) {
 61             rep(j, 0, n+1)
 62                 swap(g[r][j], g[i][j]);
 63         }
 64 
 65         rep(k, i+1, n) {
 66             if (fabs(g[i][i]) < eps)
 67                 continue;
 68             double f = g[k][i] / g[i][i];
 69             rep(j, i, n+1)
 70                 g[k][j] -= f * g[i][j];
 71         }
 72     }
 73     
 74     per(i, 0, n) {
 75         rep(j, i+1, n)
 76             g[i][n] -= g[j][n] * g[i][j];
 77         g[i][n] /= g[i][i];
 78     }
 79 }
 80 
 81 void add(int ridx, int i, int j, double val) {
 82     if (i == m) {
 83         if (j == 1)
 84             g[ridx][i*n+j-1] -= val;
 85         return ;
 86     }
 87 
 88     g[ridx][i*n+j-1] += val;
 89 }
 90 
 91 void solve() {
 92     int idx = 0;
 93 
 94     memset(g, 0, sizeof(g));
 95 
 96     rep(i, 0, m) {
 97         rep(j, 1, n+1) {
 98             add(idx, i, j, 1.0);
 99             if (j == 1) {
100                 add(idx, i+1, j, -0.25);
101                 add(idx, 1, n-2, -0.75);
102             } else if (j == 2) {
103                 add(idx, i+1, n-2, -0.25);
104                 add(idx, 1, 1, -0.25);
105                 add(idx, 1, n-1, -0.5);
106             } else if (j == 3) {
107                 add(idx, i+1, n-1, -0.25);
108                 add(idx, 1, n-1, -0.25);
109                 add(idx, 1, 1, -0.25);
110                 add(idx, 1, n, -0.25);
111             } else if (j == 4) {
112                 add(idx, i+1, n, -0.25);
113                 add(idx, 1, n, -0.5);
114                 add(idx, 1, 1, -0.25);
115             } else {
116                 add(idx, 1, j-3, -0.75);
117                 add(idx, i+1, j-3, -0.25);
118             }
119             ++idx;
120         }
121     }
122 
123     gauss_elimination(g, idx);
124     double ans = g[k-1][idx];
125     printf("%.6lf\n", ans);
126 }
127 
128 int main() {
129     ios::sync_with_stdio(false);
130     #ifndef ONLINE_JUDGE
131         freopen("data.in", "r", stdin);
132         freopen("data.out", "w", stdout);
133     #endif
134 
135     int t;
136 
137     scanf("%d", &t);
138     rep(tt, 1, t+1) {
139         scanf("%d%d%d", &n, &m, &k);
140         printf("Case #%d: ", tt);
141         solve();
142     }
143 
144     #ifndef ONLINE_JUDGE
145         printf("time = %d.\n", (int)clock());
146     #endif
147 
148     return 0;
149 }