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
时间： 2024-10-11 19:28:46