题目

分析

搜索要炸
只好DP
方程如下
$f[x][y][step]=\sum_{i=0}^{3}{f[x+dx[i]][y+dy[i]][step-1]}$f[x][y][step]=∑i=03​f[x+dx[i]][y+dy[i]][step−1]
经典建模

code

#include<bits/stdc++.h>
using namespace std;
#define loop(i,start,end) for(int i=start;i<=end;i++)
#define anti_loop(i,start,end) for(int i=start;i>=end;i--)
#define clean(arry,num); memset(arry,num,sizeof(arry));
#define min(a,b) ((a<b)?a:b)
#define ll long long
#define lowbit(a) (a&(-a))
int n;
int s_x,s_y,m_x,m_y;
const int maxn=30;
const int maxk=maxn*maxn;
int g[maxn][maxn];
int f[maxn][maxn][maxk];
int dx[4]={0,0,1,-1};
int dy[4]={-1,1,0,0};
inline int read()
{
    int neg=1;int ans=0;char r=getchar();
    while(r>'9'||r<'0'){if(r=='-')neg=-1;r=getchar();}
    while(r>='0'&&r<='9'){ans=ans*10+r-'0';r=getchar();}
    return ans*neg;
}
void DP()
{
    clean(f,0);
    f[s_x][s_y][0]=1;
    loop(k,1,n*n)
    {
        loop(x,1,n)
        {
            loop(y,1,n)
            {
                if(g[x][y]==-1)continue;
                loop(z,0,3)
                {
                    int nx=x+dx[z];
                    int ny=y+dy[z];
                    if(g[nx][ny]==-1)continue;
                    f[x][y][k]+=f[nx][ny][k-1];
                }
            }
        }
    }
    loop(k,1,n*n)//步数最大n x n，从小到大枚举，第一个不为零的即是最短路
    {
        if(f[m_x][m_y][k])
        {
            printf("%d\n%d",k,f[m_x][m_y][k]);
            return;
        }
    }
}
int main()
{
    //freopen("datain.txt","r",stdin);
    n=read();
    char s[maxn];
    clean(g,-1);
    loop(i,1,n)
    {
        scanf("%s",s);
        loop(j,0,n-1)
        {
            if(s[j]=='E'){s_x=i;s_y=j+1;g[i][j+1]=0;}
            else if(s[j]=='X'){g[i][j+1]=-1;}
            else if(s[j]=='.'){g[i][j+1]=0;}
            else if(s[j]=='S'){m_x=i;m_y=j+1;g[m_x][m_y]=0;};
        }
    }
    DP();
    return 0;
}