1536 Naval warfare

Think of a few connected blocks, ~ ~ This is a 5 points

#include <cstdio>#include<iostream>using namespacestd;Charmap[ the][ the];intn,m,ans,vis[ the][ the];voidDfsintIintj) {    if(i<1|| j<1|| i>n| | J&GT;M)return ; if(Vis[i][j])return ; if(!Vis[i][j]) {Vis[i][j]=1; if(map[i][j]=='#'){            intf1=0, f2=0, f3=0, f4=0; if((map[i+1][j]=='#'&&!vis[i+1][J]) | | map[i+1][j]!='#') f1=1; if((map[i-1][j]=='#'&&!vis[i-1][J]) | | map[i-1][j]!='#') f2=1; if((map[i][j+1]=='#'&&!vis[i][j+1])|| map[i][j+1]!='#') f3=1; if((map[i][j-1]=='#'&&!vis[i][j-1])|| map[i][j-1]!='#') f4=1; if(F1&AMP;&AMP;F2&AMP;&AMP;F3&AMP;&AMP;F4) ans++; }} dfs (i+1, J); DFS (I,j+1); DFS (i-1, J); DFS (I,j-1);}intMain () {CIN>>n>>m;  for(intI=1; i<=n;i++)         for(intj=1; j<=m;j++) Cin>>Map[i][j]; DFS (1,1); if(ANS) printf ("there is%d ships.\n", ans); Elseprintf"Bad placement.\n"); return 0;} 

Read the topic, DFS has timed out

#include <cstdio>#include<iostream>using namespacestd;Const intdx[2]={0,1};Const intdy[2]={1,0};Charmap[1010][1010];intn,m,ans,vis[1010][1010];intFlag;voidDfsintIintJintnum) {    if(i<1|| j<1|| i>n| | j>m| | Flagreturn ; if(Vis[i][j]&&vis[i][j]!=num) {flag=1;return ;} VIS[I][J]=num;  for(intk=0;k<2; k++){        intx=i+dx[k],y=j+Dy[k]; if(map[x][y]=='#'&&vis[x][y]>=0) {DFS (x,y,num); }    }}intMain () {CIN>>n>>m;  for(intI=1; i<=n;i++)         for(intj=1; j<=m;j++) Cin>>Map[i][j];  for(intI=1; i<=n&&!flag;i++){         for(intj=1; j<=m&&!flag;j++){            if(map[i][j]=='#'&&map[i+1][j]=='#'&&map[i][j+1]=='#')                if((map[i+1][j+1]=='#'|| map[i-1][j-1]=='#'|| map[i+1][j-1]=='#'|| map[i-1][j+1]=='#')); Else{flag=1; Break; }            if(map[i][j]=='#'&&!Vis[i][j]) {ans++;            DFS (I,j,ans); }        }    }    if(!flag) printf ("there is%d ships.\n", ans); Elseprintf"Bad placement.\n"); return 0;} 

To search the Internet, mostly with BFS, with a

Exercises

Although said is a silver, but I just saw when still have to have the idea Ah!!

Traverse the matrix, find an xx,yy lattice (*.) as the upper-left corner of the current rectangle, start the BFS, if the process appears x< xx| | y< yy Words is impossible (hand painting can be seen), if there is no such situation, the record x-coordinate, the y-coordinate of the maximum value of Maxx,maxy, and then record the number of lattice CNT, if cnt== (maxx-xx+1) * (maxy-yy+1) Indicates that the rectangle is full, the current rectangle conforms to the requirements, one can find it.

AC Code:

#include <queue>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespacestd;#defineN 1010Const intdx[]={-1,0,1,0};Const intdy[]={0,1,0,-1};CharMap[n][n];intvis[n][n],n,m;structnode{intx, y; Node (intx=0,inty=0): X (x), Y (y) {}};intInsideintXintY) {returnx>=1&&x<=n&&y>=1&&y<=m;}intBFsintXxintyy) {    intMaxx=xx,maxy=yy,cnt=0; Queue<node>P;    Q.push (Node (xx,yy)); VIS[XX][YY]=1;  while(!Q.empty ()) {Node K=Q.front (); Q.pop (); CNT++;  for(intI=0;i<4; i++){            intnx=k.x+dx[i],ny=k.y+Dy[i]; if(Inside (Nx,ny) &&map[nx][ny]=='#'&&!Vis[nx][ny]) {Vis[nx][ny]=1; Maxx=Max (MAXX,NX); Maxy=Max (Maxy,ny);                Q.push (Node (nx,ny)); if(nx<xx| | NY&LT;YY)return 0; }        }    }    if(cnt== (maxx-xx+1) * (maxy-yy+1))return 1; return 0;}intMain () {CIN>>n>>m; intCnt=0;  for(intI=1; i<=n;i++) scanf ("%s", map[i]+1);  for(intI=1; i<=n;i++)         for(intj=1; j<=m;j++)            if(!vis[i][j]&&map[i][j]=='#')                 if(!BFS (I,J)) {printf ("Bad placement.\n");return 0; }                Elsecnt++; printf ("there is%d ships.", CNT); return 0;}

1536 Naval warfare