地牢逃脫

是對最短路徑的變形，即是就最短路徑當中的最長路徑，注意修改即可。bfs,dfs都可以實現，現在附上dfs的實現方式。

#include <iostream>
#include <cstring>
using namespace std;
int const maxn = 100;
char data[maxn][maxn];
int r,c;
int vis[maxn][maxn];
int dr[maxn];
int dc[maxn];
int num;
int ans = -1;
void dfs(int ir,int ic) {
    for (int i=0;i<num;i++) {
        int nr = ir + dr[i];
        int nc = ic + dc[i];
        if (nr>=0&&nr<r&&nc>=0&&nc<c) {
            if(data[nr][nc]=='.') {
                if(vis[nr][nc]==-1)  {
                vis[nr][nc] = vis[ir][ic] + 1;  
                dfs(nr,nc);}
                else if (vis[nr][nc] > vis[ir][ic] + 1) {
                    vis[nr][nc] = vis[ir][ic] + 1;
                    dfs(nr,nc);
                }
            }
        }
    }
}

void f() {
for (int i=0;i<r;i++)
     for (int j=0;j<c;j++)  {
        if (data[i][j]!='.') continue;
        if (data[i][j]=='.'&&vis[i][j]==-1) {
            ans = -1;
            return ;
        } 
        if (data[i][j]=='.'&&vis[i][j]>ans) {
            ans = vis[i][j];
        }
     }  
}



int main() {
    cin>>r>>c;
    memset(vis,-1,sizeof(vis));
    for (int i=0;i<r;i++)
     for (int j=0;j<c;j++) cin>>data[i][j];
    int ir,ic;
    cin>>ir>>ic;
    cin>>num;
    for (int i=0;i<num;i++) cin>>dr[i]>>dc[i];
    vis[ir][ic]=0;
    dfs(ir,ic);
    f();
    if(ans!=-1) cout<<ans<<endl;
    else cout<<-1<<endl;
}