poj 1856Sea Battle(BFS ,DFS亦可)

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终于AC了，眼泪汪汪啊……昨天从晚上11点弄到2点，开始时TL，然后WR……今天早上再看代码，突然发现，有一个等号没写，悲催啊……

这个题的关键在于先找出‘#’围成的最大矩形，然后判断: 矩形内部是否全为‘#’; 左边界的左边，右边界的右边和下边界的下边是否全为‘.’ ;下方两个顶点的对顶点是否为‘.’;

Sea Battle

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 2429		Accepted: 832

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.

代码如下：

#include<stdio.h>
#include<string.h>
int r, c;
char map[1001][1001];
int visited[1001][1001]={0};int BFS(int x, int y)
{int iy=y, jx=x;//找出最大矩形（jx, iy）- (x, y)while(x+1<c && map[iy][x+1]=='#' )x++;while(y+1<r && map[y+1][jx]=='#' )y++;int i ,j;//先判断边界是否符合for(j=jx; j<=x; j++)//矩形的下方{if(y<r-1){if( map[y+1][j]=='#')return 0;}}for(i=iy; i<=y; i++){if(jx>=1)//矩形左边{if( map[i][jx-1]=='#')return 0;}if(x<c-1)//矩形右边{if( map[i][x+1]=='#')return 0;}}//判断矩形内部是否符合for(i=iy; i<=y; i++)for(j=jx; j<=x; j++){if(map[i][j]=='#')visited[i][j]=1;else return 0;}//再判断下方两个个顶点的对顶点if(y<r-1 && jx>=1)//左下角if( map[y+1][jx-1]=='#')return 0;if(y<r-1 && x<c-1)//右下角if( map[y+1][x+1]=='#')return 0;return 1;
}
int main()
{int i, j;int num;int flag;while(scanf("%d %d",&r,&c) && r){getchar();num=0;flag=0;for(i=0; i<r; i++)for(j=0; j<c; j++)visited[i][j]=0;for(i=0; i<r; i++){gets(map[i]);}for(i=0; i<r; i++){if( flag ) break;for(j=0; j<c; j++){if( map[i][j]=='#' && !visited[i][j]){if( BFS(j, i) )num++;else{flag=1;break;}}}}if( flag ) printf("Bad placement.\n");else  printf("There are %d ships.\n",num);}
}

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