山東省第一屆ACM大學生程式設計競賽-Balloons（搜尋）

問題及程式碼：

Balloons

Time Limit: 1000MS Memory limit: 65536K

題目描述

Both Saya and Kudo like balloons. One day, they heard that in the central park, there will be thousands of people fly balloons to pattern a big image.
They were very interested about this event, and also curious about the image.
Since there are too many balloons, it is very hard for them to compute anything they need. Can you help them?
You can assume that the image is an N*N matrix, while each element can be either balloons or blank.
Suppose element A and element B are both balloons. They are connected if:
i) They are adjacent;
ii) There is a list of element C₁,C₂, … ,C_n, whileA andC₁ are connected,C₁ andC₂ are connected …C_n and B are connected.
And a connected block means that every pair of elements in the block is connected, while any element in the block is not connected with any element out of the block.
To Saya, element A(xa,ya)and B(xb,yb) is adjacent if |xa-xb| + |ya-yb| ≤ 1
But to Kudo, element A(xa,ya) and element B (xb,yb) is adjacent if |xa-xb|≤1 and |ya-yb|≤1
They want to know that there’s how many connected blocks with there own definition of adjacent?

輸入

The input consists of several test cases.
The first line of input in each test case contains one integer N (0<N≤100), which represents the size of the matrix.
Each of the next N lines contains a string whose length is N, represents the elements of the matrix. The string only consists of 0 and 1, while 0 represents a block and 1represents balloons.
The last case is followed by a line containing one zero.

輸出

For each case, print the case number (1, 2 …) and the connected block’s numbers with Saya and Kudo’s definition. Your output format should imitate the sample output. Print a blank line after each test case.

示例輸入

示例輸出

Case 1: 3 2

提示

來源

2010年山東省第一屆ACM大學生程式設計競賽

題意：

輸入一個n行n列的矩陣，其中：0 represents a block and 1represents balloons

根據以下條件

判斷連線在一起的“1”塊有幾組。

/*
* Copyright (c) 2016, 煙臺大學計算機與控制工程學院
* All rights reserved.
* 檔名稱：dfs.cpp
* 作    者：單昕昕
* 完成日期：2016年4月1日
* 版 本 號：v1.0
*/
#include<iostream>
#include<cstring>
#include<cstdio>
#include<malloc.h>
using namespace std;
int book[101][101];
int Map[101][101];
int n,tx,ty;
int move1_x[4]= {0,1,0,-1};
int move1_y[4]= {-1,0,1,0};
int move2_x[8]= {-1,0,1,1,1,0,-1,-1};
int move2_y[8]= {-1,-1,-1,0,1,1,1,0};
void dfs1(int x,int y)
{
    int i;
    if (Map[x][y]==1&&book[x][y]==false)
    {
        book[x][y]=true;
        for(i=0; i<4; i++)
        {
            tx=x+move1_x[i];//計算座標
            ty=y+move1_y[i];
            if (tx>=0&&ty>=0&&tx<n&&ty<n&&Map[tx][ty]==1&&book[tx][ty]==false)
            {
                dfs1(tx,ty);
            }
        }
        return ;
    }
}

void dfs2(int x,int y)
{
    int i;
    if (Map[x][y]==1&&book[x][y]==false)
    {
        book[x][y]=true;
        for(i=0; i<8; i++)
        {
            tx=x+move2_x[i];//計算座標
            ty=y+move2_y[i];
            if (tx>=0&&ty>=0&&tx<n&&ty<n&&Map[tx][ty]==1&&book[tx][ty]==false)
            {
                dfs2(tx,ty);
            }
        }
        return ;
    }
}

int main()
{
    int Case=0;
    while(cin>>n&&n!=0)
    {
        memset(book,false,sizeof(book));
        int i,j,sum1=0,sum2=0;
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
                cin >> Map[i][j];
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
            {
                if(Map[i][j]==1&&book[i][j]==false)
                {
                    dfs1(i,j);
                    ++sum1;
                }
            }
        memset(book,false,sizeof(book));
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
            {
                if(Map[i][j]==1&&book[i][j]==false)
                {
                    dfs2(i,j);
                    ++sum2;
                }
            }
        cout<<"Case "<<++Case<<": "<<endl;
        cout<<sum1<<" "<<sum2<<endl;
    }
    return 0;
}
/*
5
1 1 0 0 1
0 0 1 0 0
1 1 1 1 1
1 1 0 1 0
1 0 0 1 0
*/

執行結果：