"Daily training"ice Cave (codeforces Round 301 Div.2 C)

Test instructions and analysis (Codeforces 540C)

The pit is killing me .... I was confused with a classic BFS problem, which is simpler than that one.
That is probably the case, an ice tower, the first time a block will be broken, the second crushing will fall. And then find the feasible solution.
But the question ... It's the first floor of the ice tower.
In other words, it's just a slightly limited two-dimensional maze problem.
The following is a good idea, but this data needs to be considered:

1 2XX1 11 1

The answer to this data is no. The workaround is to consider this: two arrays to record access status: Vis array and block array.

Code

#include <queue> #include <set> #include <iostream> #include <cstring> #include <algorithm > #include <vector> #define MP make_pair#define PB push_back#define fi first#define se second#define ZERO (x) Memse T ((x), 0, sizeof (x)) #define ALL (x) (x). Begin (), (x). End () #define REP (I, A, b) for (int i = (a); I <= (b); ++i) #define PE     R (I, A, b) for (int i = (a); I >= (b); i.) #define QUICKIO Ios::sync_with_stdio (FALSE);                      Cin.tie (0); Cout.tie (0); using namespace Std;bool iscracked[505][505];bool vis[505][505];set<pair<int,int> > S;int bx,    by,ex,ey,n,m;const int Dx[]={0,1,0,-1};const int Dy[]={1,0,-1,0};bool bfs () {queue<pair<int,int> > Q;    Q.push (MP (Bx,by));    S.insert (MP (Bx,by));    Iscracked[bx][by]=true;    Auto End_pair=mp (Ex,ey);        while (!q.empty ()) {Auto Now=q.front (); Q.pop (); if (Now==end_pair && iscracked[now.fi][now.se] && vis[now.fi][now.se]) return true;        cout<< "NP:" <<now.fi<< "" <<now.se<<endl;        Iscracked[now.fi][now.se]=true;        Vis[now.fi][now.se]=true;            Rep (i,0,3) {int tx=now.fi+dx[i], ty=now.se+dy[i]; if (tx>=1 && tx<=n && ty>=1 && ty<=m && (!iscracked[tx][ty]| |            (Tx==ex&&ty==ey)))                {Auto Tstat=mp (tx,ty);                cout<< "Trying to Go" <<tx<< "" <<ty<< "<<iscracked[tx][ty]<<endl; if (S.find (Tstat) ==s.end () | | tstat==end_pair) {//cout<< "Success" <<en                    dl                    S.insert (Tstat);                Q.push (Tstat); }}}} return false;}    int main () {cin>>n>>m; ZERO (iscracked);    ZERO (VIS);         Rep (i,1,n) {string str; cin>>str; Rep (j,0,m-1) {iscracked[i][j+1]=vis[i][j+1]=str[j]== ' X ';    }} cin>>bx>>by>>ex>>ey;    Vis[bx][by]=false; if (Bx==ex && by==ey && n==1 && m==1) cout<<string (BFS ()? "    YES ":" NO ") <<endl; return 0;}

Daily training "ice Cave (codeforces Round 301 Div.2 C)