算法实战1：爱消除（Elimination Game）
have you played the elimination game before? When you hit a rectangle, other rectangles which are adjacent and in same color will be cleared. We could use a n*m matrix to indicate the game board. Numbers in the matrix indicate different color. Same number means same color.
For example: this is a 5*4 game board:
1 1 2 4
2 3 3 3
2 2 3 5
3 3 3 3
1 2 3 4
The matrix index starts from 0. After you hit the rectangle at (3, 3), all the adjacent rectangle with number 3 will be cleared. We use number 0 to mark the cleared rectangle. The result will be:
1 1 2 4
2 0 0 0
2 2 0 5
0 0 0 0
1 2 0 4
Put gravity into consideration, rectangles which locates on top of the 0-rect will fall down and the result will be like this:
0 0 0 0
1 0 0 0
2 1 0 4
2 2 0 5
1 2 2 4
The question is, given a m*n matrix and a (x,y) indicate which rectangle is hit, compute the result of this hit.
Input:
4 // the number of the columns
5 // the number of the rows
1 1 2 4 // each line of the metrix. Number are separated by space.
2 3 3 3
2 2 3 5
3 3 3 3
1 2 3 4
3 // the x index of the hit position
3 // the y index of the hit position
Output:
0 0 0 0
1 0 0 0
2 1 0 4
2 2 0 5
1 2 2 4
The output should be the matrix status line by line. Within each line, the numbers are separate by space.
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这道题很适合现在火爆了的爱消除！（Eliminating Game）
分两步骤：
（1）首先点击某个数字后，希望能够删除掉颜色相近的
（2）对数组重新排列
第一步如何消除呢？我们知道，点击了某个块，我们需要遍历它的四周四个方块，如果数值相同那么也要继续遍历，
到这里很明显有两个思路：一是可以采用递归的方式，二是可以采用队列或者栈的形式来找出所有要删除的方块
第二步就是一个下沉的过程，其实就是个冒泡过程把0冒泡上去，这里要的是稳定的冒泡，不能乱了顺序
所以解法见代码咯
//============================================================================
// Name : EliminatingGame.cpp
// Author : YLF
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================
#include <iostream>
using namespace std;
//简单起见，先假设最大行数，列数20
#define ROW_NUM 20
#define COL_NUM 20
void EliminateBlock(int block[ROW_NUM][COL_NUM], int hitRow, int hitCol, int rowNum, int colNum);
void ReOrder(int block[ROW_NUM][COL_NUM], int rowNum, int colNum);
void PrintBlock(int block[ROW_NUM][COL_NUM], int rowNum, int colNum);
int main() {
int rowNum = 0;
int colNum = 0;
//input
cin>>colNum;
cin>>rowNum;
cout<<"input block"<<endl;
int block[ROW_NUM][COL_NUM];
int i = 0, j = 0;
for(i=0;i<rowNum;i++)
for(j=0;j<colNum;j++)
cin>>block[i][j];
int hitCol = 0;
int hitRow = 0;
cin>>hitCol;
cin>>hitRow;
//eliminate
EliminateBlock(block, hitRow, hitCol, rowNum, colNum);
ReOrder(block, rowNum, colNum);
PrintBlock(block, rowNum, colNum);
return 0;
}
/*
* using for eliminate
*/
void EliminateBlock(int block[ROW_NUM][COL_NUM], int hitRow, int hitCol, int rowNum, int colNum){
int value = block[hitRow][hitCol];
block[hitRow][hitCol] = 0;
//up
if(hitRow-1>=0 && block[hitRow-1][hitCol]==value)
EliminateBlock(block, hitRow-1, hitCol, rowNum, colNum);
//left
if(hitCol-1>=0 && block[hitRow][hitCol-1]==value)
EliminateBlock(block, hitRow, hitCol-1, rowNum, colNum);
//down
if(hitRow+1<rowNum && block[hitRow+1][hitCol]==value)
EliminateBlock(block, hitRow+1, hitCol, rowNum, colNum);
//right
if(hitCol+1<colNum && block[hitRow][hitCol+1]==value)
EliminateBlock(block, hitRow, hitCol+1, rowNum, colNum);
}
/*
* using for re-order
* bottom-up left-right
*/
void ReOrder(int block[ROW_NUM][COL_NUM], int rowNum, int colNum){
int j = 0;
int index1 = rowNum-1, index2 = rowNum - 1;
for(j=0;j<colNum;j++){
index1 = rowNum-1;
while(index1>=0){
while(index1>=0 && block[index1][j] != 0)
index1--;
index2 = index1-1;
while(index2>=0 && block[index2][j] == 0)
index2--;
if(index2>=0){
block[index1][j] = block[index2][j];
block[index2][j] = 0;
}
index1--;
}
}
}
/*
* print block
*/
void PrintBlock(int block[ROW_NUM][COL_NUM], int rowNum, int colNum){
int i = 0, j = 0;
cout<<endl;
for(i=0;i<rowNum;i++){
for(j=0;j<colNum;j++)
cout<<block[i][j]<<" ";
cout<<endl;
}
}
这道题给我的思考：
让我想起了之前遇到的一个面试题，考官问到类似这题的题目，当时问我除了递归还有别的非递归方法吗？
我那是绞尽脑汁，也紧张，一是半会没想出，于是的于是。。。。其实用队列是很好实现的现将我们需要删除的元素添加进去，出队列的时候把它周围值相等的加入，但是前提，加入队列后这个元素值必须置为0，否则会出现死循环了。。