集训队专题（4）1004 Where is the canteen

Where is the canteen

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1375    Accepted Submission(s): 414

Problem Description

After a long drastic struggle with himself, LL decide to go for some snack at last. But when steping out of the dormitory, he found a serious problem : he can't remember where is the canteen... Even worse is the campus is very dark at night. So, each time he move, he check front, back, left and right to see which of those four adjacent squares are free, and randomly walk to one of the free squares until landing on a canteen.

 

Input

Each case begin with two integers n and m ( n<=15,m<=15 ), which indicate the size of the campus. Then n line follow, each contain m characters to describe the map. There are 4 different type of area in the map:

'@' is the start location. There is exactly one in each case.
'#' is an impassible square.
'$' is a canteen. There may be more than one in the campus.
'.' is a free square.

 

Output

Output the expected number of moves required to reach a canteen, which accurate to 6 fractional digits. If it is impossible , output -1.

 

Sample Input

1 2@$2 2@..$1 3@#$

 

Sample Output

1.0000004.000000-1

 

题意：给出一张图，问我们从起点走到食堂所需要的步数的期望。

此题是一个比较基础的题，很明显的DP概率+高斯消元法+搜索，如果这两种方法都掌握的话解出此题并不难。

首先我们先建立dp状态方程，设步数的期望为E[x]，则E[i]=(E[next1]+E[next2]……E[nextk])/k+1，至于每个点的k为多少，我们可以通过搜索找到，在建立好所有点的状态方程之后，我们便可以开始对这个方程组进行高斯消元法求解。这里还注意可能会有多个出口。

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<cmath>using namespace std;int n,m;int dir[4][2]={0,1,0,-1,1,0,-1,0};char str[20][20];double a[230][230];bool flag[20][20];int ans;const double eps = 1e-12;double x[230];bool free_x[230];int sgn(double x){    return (x>eps)-(x<-eps);}int gauss(){    int i, j, k;    int max_r;    int col;    double temp;    int free_x_num;    int free_index;    int equ = n*m,var = n*m;    col = 0;    memset(free_x,true,sizeof(free_x));    for (k = 0; k < equ && col < var; k++, col++)    {        max_r = k;        for (i = k + 1; i < equ; i++)        {            if (sgn(fabs(a[i][col]) - fabs(a[max_r][col]))>0) max_r = i;        }        if (max_r != k)        {            for (j = k; j < var + 1; j++) swap(a[k][j], a[max_r][j]);        }        if (sgn(a[k][col]) == 0 )        {            k--; continue;        }        for (i = k + 1; i < equ; i++)        {            if (sgn(a[i][col])!=0)            {                double t = a[i][col] / a[k][col];                for (j = col; j < var + 1; j++)                {                    a[i][j] = a[i][j] - a[k][j] * t;                }            }        }    }    for(i=k;i<equ;i++)    if(sgn(a[i][col])!=0) return 0;    if (k < var)    {        for (i = k - 1; i >= 0; i--)        {            free_x_num = 0;            for (j = 0; j < var; j++)            {                if ( sgn(a[i][j])!=0 && free_x[j]){                    free_x_num++, free_index = j;                }            }            if(free_x_num>1)    continue;            temp = a[i][var];            for (j = 0; j < var; j++)            {                if (sgn(a[i][j])!=0 && j != free_index) temp -= a[i][j] * x[j];            }            x[free_index] = temp / a[i][free_index];            free_x[free_index] = 0;        }        return var - k;    }    for (i = var - 1; i >= 0; i--)    {        temp = a[i][var];        for (j = i + 1; j < var; j++)        {            if (sgn(a[i][j])!=0) temp -= a[i][j] * x[j];        }        x[i] = temp / a[i][i];    }    return 1;}void bfs(){    int s,e;    memset(flag,false,sizeof(flag));    queue<int>x,y;    while(!x.empty()) x.pop();    while(!y.empty()) y.pop();    for(int i=0;i<n;i++)    for(int j=0;j<m;j++)    if(str[i][j]=='$')    {        x.push(i);        y.push(j);        s=i;        e=j;        flag[i][j]=true;    }    while(!x.empty())    {        int ux=x.front(),uy=y.front(),vx,vy;        x.pop();y.pop();        for(int i=0;i<4;i++)        {            vx=ux+dir[i][0];            vy=uy+dir[i][1];            if(vx>=0&&vx<n&&vy>=0&&vy<m&&str[vx][vy]!='#'&&!flag[vx][vy])            {                flag[vx][vy]=true;                x.push(vx);y.push(vy);            }        }    }}int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        for(int i=0;i<n;i++)        scanf("%s",str[i]);        for(int i=0;i<=n*m;i++)        for(int j=0;j<=n*m;j++)            a[i][j]=0;        bfs();        int s,e;        for(int i=0;i<n;i++)        for(int j=0;j<m;j++)        {            int cnt=0;            if(str[i][j]=='@')            {                s=i;                e=j;            }            if(str[i][j]=='$')//目标的方程便是e[i]=0;            {                a[i*m+j][n*m]=0;                a[i*m+j][i*m+j]=1;                continue;            }            if(str[i][j]=='#') //不可达的点可以忽略                continue;            for(int k=0;k<4;k++)            {                int x=i+dir[k][0];                int y=j+dir[k][1];                if(x>=0&&x<n&&y>=0&&y<m&&str[x][y]!='#'&&flag[x][y])                {                    a[i*m+j][x*m+y]=1;                    cnt++;                }            }            a[i*m+j][n*m]=-1*cnt;            a[i*m+j][i*m+j]=-1*cnt;        }        if(flag[s][e]&&gauss())        printf("%.6f\n",x[s*m+e]);        else            puts("-1");    }    return 0;}