AtCoder Beginner Contest 109- D - Make Them Even(有意思的题)

D - Make Them Even

Time limit : 2sec / Memory limit : 1024MB

Score : 400 points

Problem Statement

There is a grid of square cells with H horizontal rows and W vertical columns. The cell at the i-th row and the j-th column will be denoted as Cell (i,j).

In Cell (i,j), aij coins are placed.

You can perform the following operation any number of times:

Operation: Choose a cell that was not chosen before and contains one or more coins, then move one of those coins to a vertically or horizontally adjacent cell.

Maximize the number of cells containing an even number of coins.

Constraints

All values in input are integers.
1≤H,W≤500
0≤aij≤9

Input

Input is given from Standard Input in the following format:

H W
a11 a12 … a1W
a21 a22 … a2W
:
aH1 aH2 … aHW

Output

Print a sequence of operations that maximizes the number of cells containing an even number of coins, in the following format:

N
y1 x1 y1' x1'
y2 x2 y2' x2'
:
yN xN yN' xN'

That is, in the first line, print an integer N between 0 and H×W (inclusive), representing the number of operations.

In the (i+1)-th line (1≤i≤N), print four integers yi,xi,yi' and xi' (1≤yi,yi'≤H and 1≤xi,xi'≤W), representing the i-th operation. These four integers represents the operation of moving one of the coins placed in Cell (yi,xi) to a vertically or horizontally adjacent cell, (yi',xi').

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Note that if the specified operation violates the specification in the problem statement or the output format is invalid, it will result in Wrong Answer.

Sample Input 1

Copy

2 3
1 2 3
0 1 1

Sample Output 1

Copy

Every cell contains an even number of coins after the following sequence of operations:

Move the coin in Cell (2,2) to Cell (2,3).
Move the coin in Cell (1,1) to Cell (1,2).

Move one of the coins in Cell (1,3) to Cell (1,2).

题解：可以从不硬币为0的网格中移动一块硬币到上下左右相邻的一个网格中，是的网格中硬币为偶数的网格尽可能的多，求移动的步数和移动的过程。这题不难但是我觉得很有意思。直接从从0,0开始遍历如果当前网格为奇数或者小于0就从接下来的网格中移动一块硬币，当达到水平最后一格下一次就从最后一格开始，跑到第一格下一次就从第一格开始，知道遍历完。用栈存储移动过程，但是最后要去除最后几个网格为-1的格子并且要使栈的大小为偶数。移动的步数就是栈的大小除2，然后在每次从栈中取两个打印就好了。

#include<bits/stdc++.h>
using namespace std;
struct  Point
{
    int x,y;
    Point(int x,int y):x(x),y(y) {}
};
int h,w,grid[1000][1000];
int main()
{
    cin>>h>>w;
    stack<Point> sta;
    int ans=0;
    for(int i=1; i<=h; i++)
    {
        for(int j=1; j<=w; j++)
            cin>>grid[i][j];
    }
    for(int i=1; i<=h; i++)
    {
        if(i%2!=0)
        {
            for(int j=1; j<=w; j++)
            {
                if(grid[i][j]%2!=0&&j<w)
                {
                    grid[i][j+1]-=1;
                    sta.push(Point(i,j));
                    sta.push(Point(i,j+1));
                }
                else if(grid[i][j]%2!=0&&j==w)
                {
                    grid[i+1][j]-=1;
                    sta.push(Point(i,j));
                    sta.push(Point(i+1,j));
                }
            }
        }
        else
        {
            for(int j=w; j>0; j--)
            {
                if(grid[i][j]%2!=0&&j>1)
                {
                    grid[i][j-1]-=1;
                    sta.push(Point(i,j));
                    sta.push(Point(i,j-1));
                }
                else if(grid[i][j]%2!=0&&j==1)
                {
                    grid[i+1][j]-=1;
                    sta.push(Point(i,j));
                    sta.push(Point(i+1,j));
                }
            }
        }
    }
    while(!sta.empty()&&grid[sta.top().x][sta.top().y]<0)
    {
        sta.pop();
    }
    if(sta.size()%2!=0)
        sta.pop();
    cout<<sta.size()/2<<endl;
    while(!sta.empty())
    {
        cout<<sta.top().x<<" "<<sta.top().y<<" ";
        sta.pop();
        cout<<sta.top().x<<" "<<sta.top().y<<endl;
        sta.pop();
    }
    return 0;
}