# Time travel

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 517    Accepted Submission(s): 80

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 !"

Sample Input
2 4 2 0 1 0 50 50 4 1 0 2 1 100

Sample Output
8.14 2.00

Source

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liuyiding

/*
HDU 4118

*/
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<queue>
#include<math.h>
using namespace std;

#define eps 1e-9
const int MAXN=220;
double a[MAXN][MAXN],x[MAXN];//方程的左边的矩阵和等式右边的值，求解之后x存的就是结果
int equ,var;//方程数和未知数个数

int Gauss()
{
int i,j,k,col,max_r;
for(k=0,col=0;k<equ&&col<var;k++,col++)
{
max_r=k;
for(i=k+1;i<equ;i++)
if(fabs(a[i][col])>fabs(a[max_r][col]))
max_r=i;
if(fabs(a[max_r][col])<eps)return 0;
if(k!=max_r)
{
for(j=col;j<var;j++)
swap(a[k][j],a[max_r][j]);
swap(x[k],x[max_r]);
}
x[k]/=a[k][col];
for(j=col+1;j<var;j++)a[k][j]/=a[k][col];
a[k][col]=1;
for(i=0;i<equ;i++)
if(i!=k)
{
x[i]-=x[k]*a[i][k];
for(j=col+1;j<var;j++)a[i][j]-=a[k][j]*a[i][col];
a[i][col]=0;
}
}
return 1;
}

int num[MAXN];
double p[MAXN];
int cnt;
int n,N;//n=2*N-2
int M;
void bfs(int s)
{
memset(num,-1,sizeof(num));
queue<int>que;
cnt=0;
num[s]=cnt++;
que.push(s);
while(!que.empty())
{
int t=que.front();
que.pop();
for(int i=1;i<=M;i++)
{
if(fabs(p[i])<eps)continue;//这点很重要，这个想到不能达到的点
int temp=(t+i)%n;
if(num[temp]==-1)
{
num[temp]=cnt++;
que.push(temp);
}
}
}
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int s,e;
int D;
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d%d%d",&N,&M,&e,&s,&D);
for(int i=1;i<=M;i++){scanf("%lf",&p[i]);p[i]/=100;}

if(e==s)//这个特判一定需要，否则可能N==1,会被0除，RE
{
printf("0.00\n");
continue;
}

n=2*(N-1);
if(D==1)s=n-s;
bfs(s);
if(num[e]==-1&&num[n-e]==-1)
{
printf("Impossible !\n");
continue;
}
equ=var=cnt;
memset(a,0,sizeof(a));
memset(x,0,sizeof(x));
for(int i=0;i<n;i++)
if(num[i]!=-1)
{
if(i==e||i==n-e)
{
a[num[i]][num[i]]=1;
x[num[i]]=0;
continue;
}
a[num[i]][num[i]]=1;
for(int j=1;j<=M;j++)
{
int t=(i+j)%n;
if(num[t]!=-1)
{
a[num[i]][num[t]]-=p[j];
x[num[i]]+=j*p[j];
}
}
}
if(Gauss())printf("%.2lf\n",x[num[s]]);
else printf("Impossible !\n");
}
return 0;
}

posted on 2012-10-06 23:09  kuangbin  阅读(1887)  评论(0编辑  收藏

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