HDU

题目：有一个n*m的字符矩阵，你要从初始位置到商店去，你每次都选择去一个可达的点（上下左右4个方向），求期望步数。

'@'初始位置，有且只有一个

'#'不能走的点

'$'商店，可能有多个

'.'可以走的点

思路：广搜以便给可达的点编号，列方程高斯消元即可

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3fint dr[]={-1,1,0,0};int dc[]={0,0,-1,1};const double _eps = 1e-12;const int MAX = 300;int equ, var;double mat[MAX][MAX], x[MAX];int sign(double x) {return (x > _eps) - (x < -_eps);}int Gauss() {int i, j, k, col, max_r, free_num, free_index;double ta, tmp;for(i = 0; i <= var; ++i) {x[i] = 0;}for(k = col = 0; k < equ && col < var; ++k, ++col) {max_r = k;for(i = k + 1; i < equ; ++i) {if(sign(fabs(mat[i][col]) - fabs(mat[max_r][col])) > 0)max_r = i;}if(max_r != k) {for(j = k; j < var + 1; ++j)swap(mat[max_r][j], mat[k][j]);}if(sign(mat[k][col]) == 0)            return 0;for(i = k + 1; i < equ; ++i) {if(sign(mat[i][col]) == 0) continue;ta = mat[i][col] / mat[k][col];for(j = col; j < var + 1; ++j)mat[i][j] -= mat[k][j] * ta;}}for(i = k; i < equ; ++i) {if(sign(mat[i][col]))return 0;}if(k < var)        return 0;for(i = var - 1; i >= 0; --i) {tmp = mat[i][var];for(j = i + 1; j < var; ++j)if(sign(mat[i][j]) != 0)tmp -= mat[i][j] * x[j];x[i] = tmp / mat[i][i];}return 1;}const int maxn=300;struct Node{    int r,c;    Node(int _r=0,int _c=0):r(_r),c(_c){}};char str[20][20];int vis[20][20],num[20][20],cnt;int n,m;bool judge(int r,int c){    if(r<0||r>=n||c<0||c>=m||str[r][c]=='#')        return false;    return true;}int bfs(int r,int c){    queue<Node> Q;    mm(vis,-1);    mm(num,0);    cnt=0;    vis[r][c]=cnt++;    Q.push(Node(r,c));    int flag=0;    while(!Q.empty()){        Node now=Q.front();        Q.pop();        if(str[now.r][now.c]=='$')            flag=1;        for(int i=0;i<4;i++){            r=now.r+dr[i];            c=now.c+dc[i];            if(!judge(r,c)) continue;            num[now.r][now.c]++;            if(vis[r][c]!=-1) continue;            vis[r][c]=cnt++;            Q.push(Node(r,c));        }    }    return flag;}void build(){    for(int i=0;i<n;i++)        for(int j=0;j<m;j++){            if(vis[i][j]==-1)                continue;         //   printf("%d\n",num[i][j]);            int pos=vis[i][j];            if(str[i][j]=='$'){                mat[pos][pos]=1;                mat[pos][cnt]=0;                continue;            }            mat[pos][pos]=1;            for(int k=0;k<4;k++){                int r=i+dr[k];                int c=j+dc[k];                if(!judge(r,c)) continue;                int t=vis[r][c];                mat[pos][t]=-1.0/num[i][j];                mat[pos][cnt]+=1.0/num[i][j];            }        }}int main(){    while(~scanf("%d%d",&n,&m)){        for(int i=0;i<n;i++)            scanf("%s",str[i]);        int r,c;        for(int i=0;i<n;i++)            for(int j=0;j<m;j++)                if(str[i][j]=='@')                    r=i,c=j;        if(!bfs(r,c)){            printf("-1\n");            continue;        }        mm(mat,0);        equ=var=cnt;        build();        if(Gauss()){            printf("%.6f\n",x[0]);        }        else{            printf("-1\n");        }    }    return 0;}