【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

 /* 4418 */
 #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 = ;
 typedef double mat[maxn][maxn];

 const double eps = 1e-;
 int n, nn, m, x, y, d;
 int visit[maxn];
 int id[maxn];
 double P[maxn];
 mat g;

 bool gauss_elimination(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]);
         }

         if (fabs(g[i][i]) < eps)
             return false;

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

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

     return true;
 }

 int bfs(int bx) {
     queue<int> Q;
     int p = ;
     bool ret = false;

     memset(visit, -, sizeof(visit));
     visit[bx] = p++;
     Q.push(bx);

     while (!Q.empty()) {
         int u = Q.front();
         Q.pop();
         if (id[u] == y)
             ret = true;
         rep(j, , m+) {
             if (fabs(P[j]) < eps)
                 continue;
             int v = (u + j) % nn;
             if (visit[v] == -) {
                 visit[v] = p++;
                 Q.push(v);
             }
         }
     }

     return ret ? p : ;
 }

 void solve() {
     if (x == y) {
         puts("0.00");
         return ;
     }

     nn = n*-;

     rep(i, , n)
         id[i] = i;
     for (int i=n,j=n-; i<nn; ++i,--j)
         id[i] = j;

     int bx;

     if (d == )
         bx = x;
     else if (d == )
         bx = nn - x;
     else
         bx = x;

     int vn = bfs(bx);
     if (!vn) {
         puts("Impossible !");
         return ;
     }

     memset(g, , sizeof(g));
     rep(i, , nn) {
         if (visit[i] == -)
             continue;

         int p = visit[i];
         g[p][p] = ;

         if (id[i] == y)
             continue;

         rep(j, , m+) {
             int k = visit[(i + j) % nn];
             if (k == -)
                 continue;

             g[p][k] -= P[j];
             g[p][vn] += j * P[j];
         }
     }

     bool flag = gauss_elimination(vn);

     if (!flag || fabs(g[][vn])<eps) {
         puts("Impossible !");
         return ;
     }

     printf("%.2lf\n", g[][vn]);
 }

 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);
     while (t--) {
         scanf("%d%d%d%d%d", &n,&m,&y,&x,&d);
         rep(i, , m+) {
             scanf("%lf", &P[i]);
             P[i] /= 100.0;
         }
         solve();
     }

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

     return ;
 }

4. 数据生成器

 import sys
 import string
 from random import randint

 def GenData(fileName):
     with open(fileName, "w") as fout:
         t = 20
         for tt in xrange(t):
             n = randint(1, 100)
             m = randint(1, 100)
             y = randint(0, n-1)
             x = randint(0, n-1)
             if x==0 or x==n-1:
                 d = -1
             else:
                 d = randint(0, 1)
             fout.write("%d %d %d %d %d\n" % (n, m, y, x, d))
             L = [0] * m
             tot = 0
             for i in xrange(m-1):
                 x = randint(0, 100-tot)
                 L[i] = x
                 tot += x
             L[m-1] = 100 - tot
             fout.write(" ".join(map(str, L)) + "\n")

 def MovData(srcFileName, desFileName):
     with open(srcFileName, "r") as fin:
         lines = fin.readlines()
     with open(desFileName, "w") as fout:
         fout.write("".join(lines))

 def CompData():
     print "comp"
     srcFileName = "F:\Qt_prj\hdoj\data.out"
     desFileName = "F:\workspace\cpp_hdoj\data.out"
     srcLines = []
     desLines = []
     with open(srcFileName, "r") as fin:
         srcLines = fin.readlines()
     with open(desFileName, "r") as fin:
         desLines = fin.readlines()
     n = min(len(srcLines), len(desLines))-1
     for i in xrange(n):
         ans2 = int(desLines[i])
         ans1 = int(srcLines[i])
         if ans1 > ans2:
             print "%d: wrong" % i

 if __name__ == "__main__":
     srcFileName = "F:\Qt_prj\hdoj\data.in"
     desFileName = "F:\workspace\cpp_hdoj\data.in"
     GenData(srcFileName)
     MovData(srcFileName, desFileName)