首页 > ACM题库 > HDU-杭电 > HDU 4418-Time travel-概率-[解题报告]HOJ
2015
07-16

HDU 4418-Time travel-概率-[解题报告]HOJ

Time travel

问题描述 :

Super Mario

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.

输入:

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.

输出:

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.

样例输入:

2
4 2 0 1 0
50 50
4 1 0 2 1
100

样例输出:

8.14
2.00

转载请注明出处,谢谢http://blog.csdn.net/acm_cxlove/article/details/7854526      
by—cxlove 

题目:给出一个数轴,有一个起点和一个终点,某个人可以走1,2,3……m步,每一种情况有一个概率,初始有一个方向,走到头则返回,问到达终点的期望步数为多少。

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

比较明显的高斯求期望问题。

E[i]表示在i点的时候到达终点的期望步数,则E[i]=E[i+1]*p1+E[i+2]*p2+……E[i+m]*pm+1。

列好方向 ,高斯消元即可。

对于这种问题,注意无解情况,只需要先BFS一次,把有效的可行状态标号,即可

#include<iostream>
#include<cstdio>
#include<map>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
#include<queue>
#define inf 1<<28
#define M 6000005
#define N 205
#define maxn 300005
#define eps 1e-8
#define zero(a) fabs(a)<eps
#define Min(a,b) ((a)<(b)?(a):(b))
#define Max(a,b) ((a)>(b)?(a):(b))
#define pb(a) push_back(a)
#define mem(a,b) memset(a,b,sizeof(a))
#define LL long long
#define lson step<<1
#define rson step<<1|1
using namespace std;
int n,m,y,x,s,d,e;
double p[N],a[N][N];
int cnt,id[N],g[N];
void debug(int n){
    for(int i=0;i<n;i++){
        for(int j=0;j<=n;j++)
           printf("%.2f ",a[i][j]);
        printf("\n");
    }
}
bool bfs(){
    queue<int>que;
    que.push(s);
    cnt=0;mem(id,-1);
    id[s]=cnt++;
    bool flag=false;
    while(!que.empty()){
        int u=que.front();que.pop();
        for(int i=1;i<=m;i++){
            int v=(u+i)%(2*n-2);
            if(zero(p[i])) continue;
            if(id[v]!=-1) continue;
            id[v]=cnt++;
            if(g[v]==y) {flag=true;}
            que.push(v);
        }
    }
    return flag;
}
void Bulid(){
    mem(a,0);
    for(int i=0;i<(2*n-2);i++){
        if(id[i]==-1) continue;
        int u=id[i];
        a[u][u]=1;
        if(g[i]==y){a[u][cnt]=0;continue;}
        for(int j=1;j<=m;j++){
            int v=(i+j)%(2*n-2);
            if(id[v]==-1) continue;
            v=id[v];
            a[u][v]-=p[j];
            a[u][cnt]+=p[j]*j;
        }
    }
}
bool gauss(int n){
    int i,j;
    for(i=0,j=0;i<n&&j<n;j++){
        int k;
        for(k=i;k<n;k++)
            if(!zero(a[k][j]))
                break;
        if(k<n){
            if(i!=k)
            for(int r=j;r<=n;r++) swap(a[i][r],a[k][r]);
            double tt=1/a[i][j];
            for(int r=j;r<=n;r++)
                a[i][r]*=tt;
            for(int r=0;r<n;r++)
                if(r!=i){
                    for(int t=n;t>=j;t--)
                        a[r][t]-=a[r][j]*a[i][t];
                }
            i++;
        }
    }
    for(int r=i;r<n;r++)
        if(!zero(a[r][n]))
            return false;
    return true;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d%d%d",&n,&m,&y,&x,&d);
        for(int i=1;i<=m;i++){scanf("%lf",&p[i]);p[i]/=100.0;}
        if(x==y){printf("0.00\n");continue;}
        if(d==0)s=x;
        else if(d==1) s=n+(n-2-x);
        else s=x;
        for(int i=0;i<n;i++) g[i]=i;
        for(int i=n,j=n-2;j>=1;j--,i++) g[i]=j;
        if(!bfs()){puts("Impossible !");continue;}
        Bulid();
        if(!gauss(cnt)) puts("Impossible !");
        else{
            for(int i=0;i<cnt;i++)
                if(zero(a[i][0]-1)){
                    double ans=a[i][cnt];
                    if(zero(ans)) printf("0.00\n");
                    else  printf("%.2f\n",ans);
                    break;
                }
        }
    }
    return 0;
}

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参考:http://blog.csdn.net/acm_cxlove/article/details/8039530