【HDOJ】4418 Time travel

1. 题目描述
K沿着$0,1,2,\cdots,n-1,n-2,n-3,\cdots,1,$的循环节不断地访问$[0, n-1]$个时光结点。某时刻，时光机故障，这导致K必须持续访问时间结点。故障发生在结点x处，方向为d，
在访问k个结点后时光机以概率$P_k%$的概率修复好，k不超过m。求当K最终访问结点Y时经过的时光结点的期望。

2. 基本思路
上述循环节包含包含$nn = 2n-2个$元素（因此，尤其需要特判n=1的情况，否则除0wa）。
通过x和方向d可以唯一的确定x在这个循环节中的位置。
设$E[i], i \in [0, nn)$表示由结点i最终访问成功Y的期望。对期望进行推导
\begin{align}
    E[0] &= (E[(0+1) \% nn]+1) \times P_1 + (E[(0+2) \% nn] + 2) \times P_2 + \cdots (E[(0+m) \% nn] + m) \times P_m \notag \\
    E[1] &= (E[(1+1) \% nn]+1) \times P_1 + (E[(1+2) \% nn] + 2) \times P_2 + \cdots (E[(1+m) \% nn] + m) \times P_m \notag \\
        &\cdots \notag \\
    E[i] &= (E[(i+1) \% nn]+1) \times P_1 + (E[(i+2) \% nn] + 2) \times P_2 + \cdots (E[(i+m) \% nn] + m) \times P_m \\
    E[i] &= 0, \quad if \quad id[i]=y
\end{align}
这里显然是一个nn元方程组，可以高斯消元解。
这题对精度有限制，因此最开始加入一个bfs，判定x能否走到y，这里一定要判定$P_j, j \in [1,m]$是否近似于0。

3. 代码

  1 /* 4418 */
  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 = 220;
 45 typedef double mat[maxn][maxn];
 46 
 47 const double eps = 1e-8;
 48 int n, nn, m, x, y, d;
 49 int visit[maxn];
 50 int id[maxn];
 51 double P[maxn];
 52 mat g;
 53 
 54 bool gauss_elimination(int n) {
 55     int r;
 56     
 57     rep(i, 0, n) {
 58         r = i;
 59         rep(j, i+1, n) {
 60             if (fabs(g[j][i]) > fabs(g[r][i]))
 61                 r = j;
 62         }
 63         
 64         if (r != i) {
 65             rep(j, 0, n+1)
 66                 swap(g[r][j], g[i][j]);
 67         }
 68         
 69         if (fabs(g[i][i]) < eps)
 70             return false;
 71         
 72         rep(k, i+1, n) {
 73             double t = g[k][i] / g[i][i];
 74             rep(j, i+1, n+1)
 75                 g[k][j] -= t * g[i][j];
 76         }
 77     }
 78     
 79     per(i, 0, n) {
 80         rep(j, i+1, n)
 81             g[i][n] -= g[i][j] * g[j][n];
 82         g[i][n] /= g[i][i];
 83     }
 84     
 85     return true;
 86 }
 87 
 88 int bfs(int bx) {
 89     queue<int> Q;
 90     int p = 0;
 91     bool ret = false;
 92     
 93     memset(visit, -1, sizeof(visit));
 94     visit[bx] = p++;
 95     Q.push(bx);
 96     
 97     while (!Q.empty()) {
 98         int u = Q.front();
 99         Q.pop();
100         if (id[u] == y)
101             ret = true;
102         rep(j, 1, m+1) {
103             if (fabs(P[j]) < eps)
104                 continue;
105             int v = (u + j) % nn;
106             if (visit[v] == -1) {
107                 visit[v] = p++;
108                 Q.push(v);
109             }
110         }
111     }
112     
113     return ret ? p : 0;
114 }
115 
116 void solve() {
117     if (x == y) {
118         puts("0.00");
119         return ;
120     }
121     
122     nn = n*2-2;
123     
124     rep(i, 0, n)
125         id[i] = i;
126     for (int i=n,j=n-2; i<nn; ++i,--j)
127         id[i] = j;
128     
129     int bx;
130     
131     if (d == 0)
132         bx = x;
133     else if (d == 1)
134         bx = nn - x;
135     else
136         bx = x;
137     
138     int vn = bfs(bx);
139     if (!vn) {
140         puts("Impossible !");
141         return ;
142     }
143     
144     memset(g, 0, sizeof(g));
145     rep(i, 0, nn) {
146         if (visit[i] == -1)
147             continue;
148         
149         int p = visit[i];
150         g[p][p] = 1;
151         
152         if (id[i] == y)
153             continue;
154         
155         rep(j, 1, m+1) {
156             int k = visit[(i + j) % nn];
157             if (k == -1)
158                 continue;
159             
160             g[p][k] -= P[j];
161             g[p][vn] += j * P[j];
162         }
163     }
164     
165     bool flag = gauss_elimination(vn);
166     
167     if (!flag || fabs(g[0][vn])<eps) {
168         puts("Impossible !");
169         return ;
170     }
171     
172     printf("%.2lf\n", g[0][vn]);
173 }
174 
175 int main() {
176     ios::sync_with_stdio(false);
177     #ifndef ONLINE_JUDGE
178         freopen("data.in", "r", stdin);
179         freopen("data.out", "w", stdout);
180     #endif
181     
182     int t;
183     
184     scanf("%d", &t);
185     while (t--) {
186         scanf("%d%d%d%d%d", &n,&m,&y,&x,&d);
187         rep(i, 1, m+1) {
188             scanf("%lf", &P[i]);
189             P[i] /= 100.0;
190         }
191         solve();
192     }
193     
194     #ifndef ONLINE_JUDGE
195         printf("time = %d.\n", (int)clock());
196     #endif
197     
198     return 0;
199 }

4. 数据生成器

 1 import sys
 2 import string
 3 from random import randint
 4 
 5     
 6 def GenData(fileName):
 7     with open(fileName, "w") as fout:
 8         t = 20
 9         for tt in xrange(t):
10             n = randint(1, 100)
11             m = randint(1, 100)
12             y = randint(0, n-1)
13             x = randint(0, n-1)
14             if x==0 or x==n-1:
15                 d = -1
16             else:
17                 d = randint(0, 1)
18             fout.write("%d %d %d %d %d\n" % (n, m, y, x, d))
19             L = [0] * m
20             tot = 0
21             for i in xrange(m-1):
22                 x = randint(0, 100-tot)
23                 L[i] = x
24                 tot += x
25             L[m-1] = 100 - tot
26             fout.write(" ".join(map(str, L)) + "\n")
27                 
28         
29 def MovData(srcFileName, desFileName):
30     with open(srcFileName, "r") as fin:
31         lines = fin.readlines()
32     with open(desFileName, "w") as fout:
33         fout.write("".join(lines))
34 
35         
36 def CompData():
37     print "comp"
38     srcFileName = "F:\Qt_prj\hdoj\data.out"
39     desFileName = "F:\workspace\cpp_hdoj\data.out"
40     srcLines = []
41     desLines = []
42     with open(srcFileName, "r") as fin:
43         srcLines = fin.readlines()
44     with open(desFileName, "r") as fin:
45         desLines = fin.readlines()
46     n = min(len(srcLines), len(desLines))-1
47     for i in xrange(n):
48         ans2 = int(desLines[i])
49         ans1 = int(srcLines[i])
50         if ans1 > ans2:
51             print "%d: wrong" % i
52 
53             
54 if __name__ == "__main__":
55     srcFileName = "F:\Qt_prj\hdoj\data.in"
56     desFileName = "F:\workspace\cpp_hdoj\data.in"
57     GenData(srcFileName)
58     MovData(srcFileName, desFileName)
59