hdu 5001 walk （概率dp+矩阵）

题目链接

Walk

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 361    Accepted Submission(s): 234
Special Judge

Problem Description

I used to think I could be anything, but now I know that I couldn't do anything. So I started traveling.

The nation looks like a connected bidirectional graph, and I am randomly walking on it. It means when I am at node i, I will travel to an adjacent node with the same probability in the next step. I will pick up the start node randomly (each node in the graph has the same probability.), and travel for d steps, noting that I may go through some nodes multiple times.

If I miss some sights at a node, it will make me unhappy. So I wonder for each node, what is the probability that my path doesn't contain it.

 

Input

The first line contains an integer T, denoting the number of the test cases.

For each test case, the first line contains 3 integers n, m and d, denoting the number of vertices, the number of edges and the number of steps respectively. Then m lines follows, each containing two integers a and b, denoting there is an edge between node a and node b.

T<=20, n<=50, n-1<=m<=n*(n-1)/2, 1<=d<=10000. There is no self-loops or multiple edges in the graph, and the graph is connected. The nodes are indexed from 1.

 

Output

For each test cases, output n lines, the i-th line containing the desired probability for the i-th node.

Your answer will be accepted if its absolute error doesn't exceed 1e-5.

 

Sample Input

   
   
   
   

    
    
    
    2
5 10 100
1 2
2 3
3 4
4 5
1 5
2 4
3 5
2 5
1 4
1 3
10 10 10
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
4 9
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0.0000000000
0.0000000000
0.0000000000
0.0000000000
0.0000000000
0.6993317967
0.5864284952
0.4440860821
0.2275896991
0.4294074591
0.4851048742
0.4896018842
0.4525044250
0.3406567483
0.6421630037
   
   
   
   

 

Source

2014 ACM/ICPC Asia Regional Anshan Online 

题意：给一个n个点，m条边的双向联通图，旅行者要旅行d 天，旅行者随机选择起点，也就是说每个点选择的概率相等。旅行者下一天会随机的去与当前点相邻的一个点（所谓相邻的点就是与该点有一条直接相连的边），求d天旅行者没有访问第K个点的概率。

题解：目标状态为d天后没有访问第K个点，用dp[ i ][ j ]表示第i天，旅行者在点j，前i-1天没有访问过点K，要到达目标状态的概率。转移为：

dp[ i ] [ j ]=sum( 1/b * dp[ w ][ j+1 ] )。b表示与点i相邻的点的个数，w为与点 i 相邻的点。我们可以用记忆化搜索的方式实现转移。这个方法可以过，但是理论复杂度偏高。

由于N只有50，正解是用矩阵实现转移。由状态转移方程，我们可以这样构造矩阵，假设图为这样时：

n=4，m=4，边如下：

1 2

2 3

3 4

2 4

构造矩阵A如下：

0 1 0 0

1/3 0 1/3 1/3

0 1/2 0 1/2

0 1/2 1/2 0

假设K等于3，构造初始矩阵B

1

1

0

1

设矩阵C为

dp [ 1 ] [ 1 ]

dp [ 1 ] [ 2 ]

dp [ 1 ] [ 3 ]

dp [ 1 ] [ 4 ]

那么C=B*（A）^d。

复杂度为：50*50^3*logd。

矩阵乘法的代码如下：

#include<stdio.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<iostream>
#include<string.h>
#include<string>
#include<math.h>
#include<stdlib.h>
#define inff 0x3fffffff
#define eps 1e-8
#define nn 55
#define mod 1000000007
typedef long long LL;
using namespace std;
int n,m,d;
struct node
{
    int en,next;
}E[nn*nn];
int p[nn],num;
void init()
{
    memset(p,-1,sizeof(p));
    num=0;
}
void add(int st,int en)
{
    E[num].en=en;
    E[num].next=p[st];
    p[st]=num++;
}
struct Matrix
{
    int l,r;
    double a[55][55];
    void clear()
    {
        l=r=0;
        memset(a,0,sizeof(a));
    }
    Matrix operator*(const Matrix &b) const
    {
        Matrix tem;
        tem.clear();
        tem.l=l,tem.r=b.r;
        for(int i=1;i<=tem.l;i++)
        {
            for(int j=1;j<=tem.r;j++)
            {
                for(int k=1;k<=r;k++)
                    tem.a[i][j]+=a[i][k]*b.a[k][j];
            }
        }
        return tem;
    }
}ma,ix,ans;
Matrix operator^(Matrix b,int y)
{
    Matrix tem;
    tem.clear();
    tem.l=b.l,tem.r=b.r;
    for(int i=1;i<=tem.l;i++)
    {
        for(int j=1;j<=tem.r;j++)
        {
            if(i==j)
                tem.a[i][j]=1;
        }
    }
    while(y)
    {
        if(y%2)
        {
            tem=tem*b;
        }
        y/=2;
        b=b*b;
    }
    return tem;
}
int main()
{
    int t,i,u,v,j,k,w;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&d);
        init();
        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&u,&v);
            add(u,v);
            add(v,u);
        }
        int bian;
        ma.clear();
        ma.l=ma.r=n;
        for(i=1;i<=n;i++)
        {
            bian=0;
            for(j=p[i];j+1;j=E[j].next)
            {
                bian++;
            }
            for(j=p[i];j+1;j=E[j].next)
            {
                w=E[j].en;
                ma.a[i][w]=(1.0)/bian;
            }
        }
        for(i=1;i<=n;i++)
        {
            ix.clear();
            ix.l=ix.r=n;
            for(j=1;j<=n;j++)
            {
                for(k=1;k<=n;k++)
                {
                    if(j==i)
                        ix.a[j][k]=0;
                    else
                        ix.a[j][k]=ma.a[j][k];
                }
            }
            ans.clear();
            ans.l=n,ans.r=1;
            for(j=1;j<=n;j++)
            {
                if(i==j)
                    ans.a[j][1]=0;
                else
                    ans.a[j][1]=(1.0/n);
            }
            ans=(ix^d)*ans;
            double re=0;
            for(j=1;j<=n;j++)
                re+=ans.a[j][1];
            printf("%.8lf\n",re);
        }
    }
    return 0;
}

记忆化搜索代码如下：

#include<stdio.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<iostream>
#include<string.h>
#include<string>
#include<math.h>
#include<stdlib.h>
#define inff 0x3fffffff
#define eps 1e-8
#define nn 55
#define mod 1000000007
typedef long long LL;
using namespace std;
int n,m,d;
struct node
{
    int en,next;
}E[nn*nn];
int p[nn],num;
void init()
{
    memset(p,-1,sizeof(p));
    num=0;
}
void add(int st,int en)
{
    E[num].en=en;
    E[num].next=p[st];
    p[st]=num++;
}
double dp[55][11000];
bool vis[55][11000];
int jin;
double dfs(int id,int step)
{
    if(id==jin)
        return 0;
    if(vis[id][step])
        return dp[id][step];
    if(step==d)
        return 1.0;
    vis[id][step]=true;
    int i,w;
    dp[id][step]=0;
    int bian=0;
    double ix=0;
    for(i=p[id];i+1;i=E[i].next)
    {
        w=E[i].en;
        ix+=dfs(w,step+1);
        bian++;
    }
    dp[id][step]=ix/bian;
    return dp[id][step];
}
int main()
{
    int t,i,u,v,j,k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&d);
        init();
        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&u,&v);
            add(u,v);
            add(v,u);
        }
        double ans;
        for(i=1;i<=n;i++)
        {
            ans=0;
            for(j=1;j<=n;j++)
            {
                for(k=0;k<=d;k++)
                    vis[j][k]=false;
            }
            jin=i;
            for(j=1;j<=n;j++)
            {
                ans+=(1.0/n)*dfs(j,0);
            }
            printf("%.8lf\n",ans);
        }
    }
    return 0;
}