POJ 1856 Sea Battle（dfs）

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Description

During the Summit, the armed forces will be highly active. The police will monitor Prague streets, the army will guard buildings, the Czech air space will be full of American F-16s. Moreover, the ships and battle cruisers will be sent to guard the banks of the Vltava river. Unfortunately, in the case of any incident, the Czech Admiralty have only a few captains able to control over the large sea battle. Therefore, it was decided to educate new admirals. As an excellent preparation, the game of "Sea Battle" was chosen to help with their study program. 

In this well-known game, a predefined number of ships of predefined shapes are placed on the square board in such a way that they cannot contact one another even with their corners. In this task, we will consider rectangular shaped ships only. The unknown number of rectangular ships of unknown sizes are placed on a rectangular board. All the ships are full rectangles built of hash characters. Write a program that counts the total number of ships present in the field. 

Input

The input consists of more scenarios. The description of each scenario begins with two integer numbers R and C separated with a single space, 1 <= R,C <= 1000. These numbers give the number of rows and columns in the game field. 

After these two numbers, there are R lines, each of them containing C characters. Each character is either hash ("#") or dot ("."). Hashes denote ships, dots water. 

Then, the next scenario description begins. At the end of the input, there will be a line containing two zeros instead of the field size. 

Output

Output a single line for every scenario. If the ships were placed correctly (i.e., there are only rectangles that do not touch each other even with a corner), print the sentence "There are S ships." where S is the number of ships. 

Otherwise, print the sentence "Bad placement.". 

Sample Input

6 6.....###...###...#..#..#.....#######6 8.....#.###.....###.....#.......##......##..#...#0 0

Sample Output

Bad placement.There are 5 ships.

Source

CTU Open 2002

                                                                                                                                                                             #

題意：就是#必須是矩陣，即為長方形或正方形，還有每個#矩形的角不能共著，如第一個圖中的##  因為他們有一個點共著了，而題目問的是只要有供著的，就輸出

Bad placement.

   

否則輸出矩形個數

比賽時沒有好的思路，糾結好久，後來看別人代碼，碉堡了，每次看到#，就dfs,記錄搜到的最大x，y，最小x，y，和#的個數  temp  ，if(temp==(maxx-minx+1)*(maxy-miny+1))，那麼就是滿足條件的，否者不滿足輸出

Bad placement.

不說了，上代碼了

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 1005char a[N][N];int n,m;int minx,maxx,miny,maxy;int temp;void dfs(int x,int y){    a[x][y]='.';    temp++;    int i,j;    minx=min(minx,x);    maxx=max(maxx,x);    miny=min(miny,y);    maxy=max(maxy,y);    for(i=-1;i<=1;i++)        for(j=-1;j<=1;j++)    {        int xx=x+i;        int yy=y+j;        if(xx>=0&&xx<n&&yy>=0&&yy<m&&a[xx][yy]=='#')            dfs(xx,yy);    }}int main(){    int i,j;    while(scanf("%d%d",&n,&m),n+m)    {        for(i=0;i<n;i++)            scanf("%s",a[i]);        int ans=0,flag=0;        for(i=0;i<n;i++)            for(j=0;j<m;j++)            if(a[i][j]=='#')            {                minx=maxx=i;                miny=maxy=j;                temp=0;                dfs(i,j);                if(temp==(maxx-minx+1)*(maxy-miny+1))                   ans++;                   else                {                    flag=1;                    i=n;                    j=m;                }            }            if(flag)                printf("Bad placement.\n");            else                printf("There are %d ships.\n",ans);    }    return 0;}