Monday, March 13, 2017

Solution of Light OJ 1066 - Gathering Food

/*  Bismillahir Rahmanir Rahim
    Solution-Using BFS
    Apply BFS from every letter(not the last) orderly
*/
#include<bits/stdc++.h>
#define fi(n, m) for(int i=n; i<=m; i++)
#define fii(i, n, m) for(int i=n; i<=m; i++)
using namespace std;
vector<int>vt[130];
int sts[130], ar[27], cnt[130], vis[130], c, ans, n, fl;

void bfs(int st, int en){
    queue<int>q;
    q.push(st);
    fii(ii, 0, n*n) {vis[ii]=0, cnt[ii]=0;}
    vis[st]=1;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int j=0; j<vt[u].size(); j++){
            int v=vt[u][j];
//if(not visited && not '#' && not getter than the next letter)
            if(!vis[v]&&sts[v]!=-1&&sts[v]<=en){
                if(sts[v]==en){ //if(find the letter)
                    c=cnt[u]+1; return ;
                }
                else cnt[v]=cnt[u]+1;
                q.push(v);
                vis[v]=1;
            }
        }
    }
}

int main(){
    int t, cs=1, cnt, m;
    cin>>t;
    while(t--){
        cin>>n;
        cnt=0, ans=0, m=0;
        fii(i, 1, n){
            fii(j, 1, n){
                char ch;
                cin>>ch;
                cnt++;
                if(ch=='.') sts[cnt]=0;
                else if(ch=='#') sts[cnt]=-1;
                else{
                    int x=(ch-'A')+1;
                    m=max(m, x); // highest letter
                    sts[cnt]=x;
                    ar[x]=cnt; //saving the position of letters
                }
                if(i>1){
                    vt[cnt].push_back(cnt-n);
                    vt[cnt-n].push_back(cnt);
                }
                if(j>1){
                    vt[cnt].push_back(cnt-1);
                    vt[cnt-1].push_back(cnt);
                }
            }
        }
        /* Input End */
        cout<<"Case "<<cs++<<": ";
        fi(1, m-1){
            c=0, fl=0;
            bfs(ar[i], i+1); // BFS for every letter(not last)
            if(c==0){  // if(fail to get next letter)
                fl=1; break;
            }
            else ans=ans+c;
        }
        if(fl) cout<<"Impossible"<<endl;
        else cout<<ans<<endl;
        fi(0, n*n) vt[i].clear();
    }
    return 0;
}

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