HDU 2612-find A-on-the-it (detail BFS)

Topic Link: Find A-to

The topic is not difficult, a few days ago, at that time prepared to write two-way BFS, and then deal with the details of some problems, to catch up with something on hold. Rewrite this evening, no two-way, with two times BFS search, and two-way BFS almost the same principle. It's just a little hole in the problem, you need to be careful.

1.Y cannot go through M. M cannot go through Y. In other words, Y and M are squares that can be thought of as walls.

2. KFC must be both Y and M to reach, only one of them arrives.

The ratio is the following column data; The answer is neither 22 nor 88 but 110, the KFC full condition in the lower left corner

5 5Y.. #@... m.....#.....@ ....

A little bit of a hole for me.

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Thank you nickname: Zstu_jayye Jay did not read his message in the discuss board. I don't expect it .

31MS

852K

Code is very slag, rushing.

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#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <queue >const int N = 1e6;const int M = 220;using namespace Std;char mapp[m][m];bool vis1[m][m],vis2[m][m];int Dis1[M][M],dis2 M  [M],n,m,l;struct node{int X,y,a;node () {x = 0; y = 0;a = 0; }};struct node{int x, y;}    Kfc[40010];int mv[4][2]={{-1,0},{1,0},{0,-1},{0,1}};void BFS (int x,int y,bool visit[m][m],int Ans[M][M]) {node f,t;    queue<node>q;    Visit[x][y]=true; F.A = 0;f.x=x;f.y=y;q.push (f), while (!q.empty ()) {t = Q.front (); Q.pop (); if (mapp[t.x][t.y]== ' @ ') ans[t. X][T.Y] = T.A; for (int i=0;i<4;i++) {f.x=t.x+mv[i][0]; f.y=t.y+mv[i][1]; if (!visit[f.x][f.y]&&0<=f.x&&f.x <n&&0<=f.y&&f.y<m&&mapp[f.x][f.y]!= ' # ') {F.A = t.a + 1; visit[f.x][f.y]=true; q.push (f);           }}}}int Main () {int Sx,sy,ex,ey;int minn;while (~scanf ("%d%d%*c", &n,&m)) {L = 0;for (int i=0;i<n;i++) {  scanf ("%s", Mapp[i]);  for (int j=0;j<m;j++) {if (mapp[i][j]== ' Y ') {sx=i; sy=j;mapp[i][j]= ' # ';}  else if (mapp[i][j]== ' M ') {mapp[i][j]= ' # '; ex=i; Ey=j;}            else if (mapp[i][j]== ' @ ') {kfc[l].x = i; kfc[l++].y = j;     }}}memset (Vis1,0,sizeof (VIS1));        memset (dis1,0,sizeof (DIS1));     BFS (SX,SY,VIS1,DIS1); memset (Vis2,0,sizeof (VIS2)); memset (dis2,0,sizeof (DIS2)); BFS (EX,EY,VIS2,DIS2); minn=n;for (int i = 0;i<l;i++) {if (Minn > DIS1[KFC[I].X][KFC[I].Y] + DIS2[KFC [I].X][KFC[I].Y]) if (DIS1[KFC[I].X][KFC[I].Y]!=0&AMP;&AMP;DIS2[KFC[I].X][KFC[I].Y]!=0) Minn        = Dis1[kfc[i].x][kfc[i].y] + dis2[kfc[i].x][kfc[i].y]; }printf ("%d\n", minn*11);} return 0;}

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HDU 2612-find A-on-the-it (detail BFS)