【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]。
\[
dp[m][j] = \begin{cases}
\begin{aligned}
&1, j=1 \\
&0, j>1
\end{aligned}
\end{cases}
\]
$dp[i][j] =$
\[
\quad \left\{ \begin{aligned}
&\frac{1}{4}dp[i+1][j] + \frac{3}{4}dp[1][n-2], &j=1 \\
&\frac{1}{4}dp[i+1][n-2] + \frac{1}{4}dp[1][1] + \frac{2}{4}dp[1][n-1], &j=2 \\
&\frac{1}{4}dp[i+1][n-3] + \frac{1}{4}dp[1][n-1] + \frac{1}{4}dp[1][1] + \frac{1}{4}dp[1]
, &j=3 \\
&\frac{1}{4}dp[i+1]
+ \frac{2}{4}dp[1]
+ \frac{1}{4}dp[1][1], &j=4 \\
&\frac{3}{4}dp[1][j-3] + \frac{1}{4}dp[i+1][j-3], &j>4
\end{aligned}
\right .
\]
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 = 105;
const double eps = 1e-8;
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, 0, n) {
r = i;
rep(j, i+1, n) {
if (fabs(g[j][i]) > fabs(g[r][i]))
r = j;
}
if (r != i) {
rep(j, 0, n+1)
swap(g[r][j], g[i][j]);
}

rep(k, i+1, n) {
if (fabs(g[i][i]) < eps)
continue;
double f = g[k][i] / g[i][i];
rep(j, i, n+1)
g[k][j] -= f * g[i][j];
}
}

per(i, 0, n) {
rep(j, i+1, n)
g[i]
-= g[j]
* g[i][j];
g[i]
/= g[i][i];
}
}

void add(int ridx, int i, int j, double val) {
if (i == m) {
if (j == 1)
g[ridx][i*n+j-1] -= val;
return ;
}

g[ridx][i*n+j-1] += val;
}

void solve() {
int idx = 0;

memset(g, 0, sizeof(g));

rep(i, 0, m) {
rep(j, 1, n+1) {
add(idx, i, j, 1.0);
if (j == 1) {
add(idx, i+1, j, -0.25);
add(idx, 1, n-2, -0.75);
} else if (j == 2) {
add(idx, i+1, n-2, -0.25);
add(idx, 1, 1, -0.25);
add(idx, 1, n-1, -0.5);
} else if (j == 3) {
add(idx, i+1, n-1, -0.25);
add(idx, 1, n-1, -0.25);
add(idx, 1, 1, -0.25);
add(idx, 1, n, -0.25);
} else if (j == 4) {
add(idx, i+1, n, -0.25);
add(idx, 1, n, -0.5);
add(idx, 1, 1, -0.25);
} else {
add(idx, 1, j-3, -0.75);
add(idx, i+1, j-3, -0.25);
}
++idx;
}
}

gauss_elimination(g, idx);
double ans = g[k-1][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, 1, t+1) {
scanf("%d%d%d", &n, &m, &k);
printf("Case #%d: ", tt);
solve();
}

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

return 0;
}