HDU 4418 Time travel

Time travel

http://acm.hdu.edu.cn/showproblem.php?pid=4418

分析：

　　因为走到最后在折返，可以将区间复制一份，就变成了只往右走，012343210。

　　写出转移方程：

$f[t] = 0$

$f[i] = p_1 \times (f[i +1] + 1) + p_2 \times (f[i +2] + 2) + \cdots$

$ =\sum\limits_{j=1}^{m}p_j \times (f[i + j] + j) $

$= \sum\limits_{j=1}^{m}p_j \times f[i + j] + \sum\limits_{j=1}^{m}p_j\times j$

然后列出线性方程组，用高斯消元求解。

$f[i] = \sum\limits_{j=1}^{m}p_j \times f[i + j] + \sum\limits_{j=1}^{m}p_j\times j$

$\sum\limits_{j=1}^{m}p_j\times j = f[i] - \sum\limits_{j=1}^{m}p_j \times f[i + j]$

补上其他项就是：

$sum =0 \times f[0] +0 \times f[1] + \cdots + f[i] + p_1 \times f[i+1] + p_2 \times f[i+2] + \cdots +0 \times f[n-1] + 0 \times f[n]$

首先bfs判一下能不能到达这个点。

代码：

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<iostream>
 #include<cctype>
 #include<set>
 #include<vector>
 #include<queue>
 #include<map>
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }
 const int N = ;
 const double eps = 1e-;
 double A[N][N], p[N], sum;
 bool vis[N];
 int n, m, s, t;
 int q[N];

 bool bfs() {
     memset(vis, false, sizeof(vis));
     int L = , R = ;
     vis[s] = ;
     q[++R] = s;
     while (L <= R) {
         int u = q[L ++];
         for (int i=; i<=m; ++i) {
             int v = (u + i) % n;
             if (!vis[v] && fabs(p[i]) > eps) vis[v] = , q[++R] = v;
         }
     }
     return vis[t] || vis[n - t]; // n-t折叠后的另一侧的点
 }
 void build() {
     memset(A, , sizeof(A));
     for (int i=; i<n; ++i) {
         A[i][i] += ;
         if (!vis[i]) { A[i][n] = 1e9; continue;}
         if (i == t || i == n - t) { A[i][n] = ; continue; }
         A[i][n] = sum;
         for (int j=; j<=m; ++j)
             A[i][(i + j) % n] -= p[j];
     }
 }
 void Gauss() {
     for (int k=; k<n; ++k) {
         int r = k;
         for (int i=k+; i<n; ++i)
             if (fabs(A[i][k]) > fabs(A[r][k])) r = i;
         if (k != r) for (int j=k; j<n; ++j) swap(A[k][j], A[r][j]);
         for (int i=k+; i<n; ++i) {
             if (fabs(A[i][k]) > eps) {
                 double t = A[i][k] / A[k][k];
                 for (int j=k+; j<=n; ++j) A[i][j] -= t * A[k][j]; //小于等于n
             }
         }
     }
     for (int i=n-; i>=; --i) {
         for (int j=i+; j<=n; ++j)
             A[i][n] -= A[j][n] * A[i][j];
         A[i][n] /= A[i][i];
     }
     printf("%.2lf\n",A[s][n]);
 }
 int main() {
     int T, d;
     scanf("%d",&T);
     while (T--) {
         scanf("%d%d%d%d%d",&n, &m, &t, &s, &d);
         n = (n - ) << ;
         sum = ;
         for (int i=; i<=m; ++i) {
             scanf("%lf",&p[i]);
             p[i] = p[i] / 100.0;
             sum += p[i] * i;
         }
         if (s == t) { puts("0.00");continue; }
         if (d) s = (n - s) % n;
         if (!bfs()) { puts("Impossible !"); continue;}
         build();
         Gauss();
     }
     return ;
 }