Time travel
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1739 Accepted Submission(s): 394
Problem Description
Agent K is one of the greatest agents in a secret organization called Men in Black. Once he needs to finish a mission by traveling through time with the Time machine. The Time machine can take agent K to some point (0 to n-1) on the timeline and when he gets
to the end of the time line he will come back (For example, there are 4 time points, agent K will go in this way 0, 1, 2, 3, 2, 1, 0, 1, 2, 3, 2, 1, ...). But when agent K gets into the Time machine he finds it has broken, which make the Time machine can‘t
stop (Damn it!). Fortunately, the time machine may get recovery and stop for a few minutes when agent K arrives at a time point, if the time point he just arrive is his destination, he‘ll go and finish his mission, or the Time machine will break again. The
Time machine has probability Pk% to recover after passing k time points and k can be no more than M. We guarantee the sum of Pk is 100 (Sum(Pk) (1 <= k <= M)==100). Now we know agent K will appear at the point X(D is the direction of the Time machine: 0 represents
going from the start of the timeline to the end, on the contrary 1 represents going from the end. If x is the start or the end point of the time line D will be -1. Agent K want to know the expectation of the amount of the time point he need to pass before
he arrive at the point Y to finish his mission.
If finishing his mission is impossible output "Impossible !" (no quotes )instead.
Input
There is an integer T (T <= 20) indicating the cases you have to solve. The first line of each test case are five integers N, M, Y, X .D (0< N,M <= 100, 0 <=X ,Y < 100 ). The following M non-negative integers represent Pk in percentile.
Output
For each possible scenario, output a floating number with 2 digits after decimal point
If finishing his mission is impossible output one line "Impossible !"
(no quotes )instead.
Sample Input
2 4 2 0 1 0 50 50 4 1 0 2 1 100
Sample Output
8.14 2.00
Source
2012 ACM/ICPC Asia Regional Hangzhou Online
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Statistic | Submit | Discuss
| Note
很明显的概率ｄｐ，设dp[i]表示在点ｉ的时候达到终点的期望步数
dp[i] = p1 * (dp[i + 1] + 1) + p2 * (dp[i + 2] + 2) + ...+ pm * (dp[i + m] + m);
当然 dp[e] = 0;
但是由于方向有２个，所以我们要把反着走的状态变成正着走，也就是把路对称一下， ０１２３４５ －》 ０１２３４５６７８９
这样 比如 ４ 走到 ３就变成了６走到７，方向就固定了，然后就可以建立线性方程组求解了，注意有些点到不了，这个得先用ｂｆｓ处理一下，顺便给每一个可到的点编号，然后高斯消元就行了
/*************************************************************************
> File Name: hdu4418.cpp
> Author: ALex
> Mail: [email protected]
> Created Time: 2014年12月29日 星期一 17时01分42秒
************************************************************************/
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 210;
const double eps = 1e-12;
double sum_p;
double mat[N][N], x[N];
double p[N];
bool free_x[N];
int ord[N];
int n, m, s, e, cnt;
int Gauss()
{
int col, k, equ, var, max_r;
int free_index, free_num;
equ = cnt;
var = cnt;
k = col = 0;
for (int i = 0; i < var; ++i)
{
free_x[i] = true;
x[i] = 0;
}
for (; k < equ && col < var; ++k, ++col)
{
max_r = k;
for (int i = k + 1; i < equ; ++i)
{
if (fabs(mat[i][col]) - fabs(mat[max_r][col]) > eps)
{
max_r = i;
}
}
if (max_r != k)
{
for (int j = k; j < var + 1; ++j)
{
swap(mat[k][j], mat[max_r][j]);
}
}
if (fabs(mat[k][col]) <= eps)
{
--k;
continue;
}
for (int i = k + 1; i < equ; ++i)
{
double tmp = mat[i][col] / mat[k][col];
for (int j = col; j < var + 1; ++j)
{
mat[i][j] -= tmp * mat[k][j];
}
}
}
//判断解的情况
for (int i = k; i < equ; ++i)
{
if (fabs(mat[i][var]) > eps)
{
return 0;
}
}
if (k < var)
{
for (int i = k - 1; i >= 0; --i)
{
free_num = 0;
for (int j = 0; j < var; ++j)
{
if (fabs(mat[i][j]) > eps && free_x[j])
{
free_num++;
free_index = j;
}
}
if (free_num > 1)
{
continue;
}
double tmp = mat[i][var];
for (int j = 0; j < var; ++j)
{
if (fabs(mat[i][j]) > eps && j != free_index)
{
tmp -= mat[i][j] * x[j];
}
}
free_x[free_index] = false;
x[free_index] = tmp / mat[i][free_index];
}
return var - k;
}
for (int i = var - 1; i >= 0; --i)
{
double tmp = mat[i][var];
for (int j = i + 1; j < var; ++j)
{
if (fabs(mat[i][j]) > eps)
{
tmp -= x[j] * mat[i][j];
}
}
x[i] = tmp / mat[i][i];
}
return 1;
}
void bfs()
{
cnt = 0;
queue <int> qu;
memset (ord, -1, sizeof(ord));
ord[s] = cnt++;
while (!qu.empty())
{
qu.pop();
}
qu.push(s);
while (!qu.empty())
{
int u = qu.front();
qu.pop();
for (int i = 1; i <= m; ++i)
{
if (p[i] <= eps)
{
continue;
}
int v = (i + u) % n;
if (ord[v] == -1)
{
ord[v] = cnt++;
qu.push(v);
}
}
}
}
void build()
{
for (int i = 0; i < n; ++i)
{
if (ord[i] == -1)
{
continue;
}
if (i == e || i == (n - e) % n)
{
mat[ord[i]][ord[i]] = 1;
mat[ord[i]][cnt] = 0;
continue;
}
mat[ord[i]][ord[i]] = 1.0;
for (int j = 1; j <= m; ++j)
{
if (ord[(i + j) % n] == -1)
{
continue;
}
mat[ord[i]][ord[(i + j) % n]] -= p[j];
}
mat[ord[i]][cnt] = sum_p;
}
}
int main()
{
int t, D;
scanf("%d", &t);
while (t--)
{
sum_p = 0;
memset (mat, 0, sizeof(mat));
scanf("%d%d%d%d%d", &n, &m, &e, &s, &D);
n = 2 * n - 2;
for (int i = 1; i <= m; ++i)
{
scanf("%lf", &p[i]);
p[i] /= 100.0;
sum_p += i * p[i];
}
if (s == e || n == 1)
{
printf("0.00\n");
continue;
}
if (D > 0)
{
s = (n - s) % n;
}
bfs();
if (ord[e] == -1 && ord[(n - e) % n] == -1)
{
printf("Impossible !\n");
continue;
}
build();
if (!Gauss())
{
printf("Impossible !\n");
continue;
}
printf("%.2f\n", x[ord[s]]);
}
return 0;
}