Codeforces 590C: (BFS)

The road has made three countries Unicom, asking at least how many grids to build roads

Enumerate each lattice, calculate the shortest possible three countries to reach this lattice, take the smallest

Find pair used instead of node and sometimes it works.

#include"Cstdio"#include"Queue"#include"Cmath"#include"Stack"#include"iostream"#include"algorithm"#include"CString"#include"Map"#include"Queue"#include"Vector"#definell Long Longusing namespacestd;Const intMAXN =1050;Const intMaxe =400050;Const intINF =1e8;CharMAT[MAXN][MAXN];intdis[3][MAXN][MAXN];intdx[4]={-1,1,0,0};intdy[4]={0,0,1,-1};intn,m;BOOLingintIintj) {    if(i<0|| I>=n)return false; if(j<0|| J&GT;=M)return false; return true;}voidSolveChara) {Queue<pair<int,int> >Q;  while(!q.empty ()) Q.pop ();  for(intI=0; i<n;i++)     for(intj=0; j<m;j++)if(mat[i][j]==a) {Q.push (pair<int,int>(i,j)); Dis[a-'1'][i][j]=0; }     while(!Q.empty ()) {Pair<int,int> t=Q.front (); Q.pop ();  for(intI=0;i<4; i++){            inttx=t.first+Dx[i]; intTy=t.second+Dy[i]; if(!ing (tx,ty) | | mat[tx][ty]=='#')Continue; if(dis[a-'1'][tx][ty]>dis[a-'1'][t.first][t.second]+ (mat[t.first][t.second]=='.') ) {Dis[a-'1'][tx][ty]=dis[a-'1'][t.first][t.second]+ (mat[t.first][t.second]=='.'); Q.push (Pair<int,int>(Tx,ty)); }        }    }}intMain () {scanf ("%d%d",&n,&m);  for(intk=0;k<3; k++)     for(intI=0; i<n;i++)     for(intj=0; j<m;j++) dis[k][i][j]=INF;  for(intI=0; i<n;i++) scanf ("%s", Mat[i]);  for(intI=0;i<3; i++) Solve (i+'1'); intans=inf*3; //solve (' 1 ');     for(intI=0; i<n;i++)     for(intj=0; j<m;j++) Ans=min (ans,dis[0][i][j]+dis[1][i][j]+dis[2][i][j]+ (mat[i][j]=='.')); if(Ans>=inf) printf ("-1\n"); Elseprintf"%d\n", ans); return 0;}

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Codeforces 590C: (BFS)