poj3279(dfs+二进制枚举思路)

Posted 2020-10-26 萧魂不散

题意转载自https://www.cnblogs.com/blumia/p/poj3279.html

题目属性：DFS

相关题目：poj3276

题目原文：
【desc】
Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M × N grid (1 ≤ M ≤ 15; 1 ≤ N ≤ 15) of square tiles, each of which is colored black on one side and white on the other side.

As one would guess, when a single white tile is flipped, it changes to black; when a single black tile is flipped, it changes to white. The cows are rewarded when they flip the tiles so that each tile has the white side face up. However, the cows have rather large hooves and when they try to flip a certain tile, they also flip all the adjacent tiles (tiles that share a full edge with the flipped tile). Since the flips are tiring, the cows want to minimize the number of flips they have to make.

Help the cows determine the minimum number of flips required, and the locations to flip to achieve that minimum. If there are multiple ways to achieve the task with the minimum amount of flips, return the one with the least lexicographical ordering in the output when considered as a string. If the task is impossible, print one line with the word "IMPOSSIBLE".
【In】
Line 1: Two space-separated integers: M and N
Lines 2.. M+1: Line i+1 describes the colors (left to right) of row i of the grid with N space-separated integers which are 1 for black and 0 for white 
【Out】
Lines 1.. M: Each line contains N space-separated integers, each specifying how many times to flip that particular location. 
【SampIn】
4 4
1 0 0 1
0 1 1 0
0 1 1 0
1 0 0 1
【SampOut】
0 0 0 0
1 0 0 1
1 0 0 1
0 0 0 0

【一句话题意】

给定长宽的黑白棋棋盘摆满棋子，每次操作可以反转一个位置和其上下左右共五个位置的棋子的颜色，求要使用最少翻转次数将所有棋子反转为黑色所需翻转的是哪些棋子（这句话好长...）。

【题目分析】

盯着奇怪的题目看了半天发现和奶牛没什么卵关系..奶牛智商高了产奶高是什么鬼说法...

这题刚开始被放到搜索的分类下了..然而这和搜索有什么关系..没经验所以想了各种和搜索沾边的方法，结果没想出解法，直到看了网上的题解，压根不是搜索啊有木有。

然后这题网上给的分类是简单题..简单题..蒟蒻跪了..

好了开始说解法。首先根据题目，每次操作都会影响到周围的“棋子”，而要使得每个1都被反转为0，那么我们就应当每次都反转1下方的棋子以改变1为0.

那么，当我们处理过1到n-1行的时候，前n-1行就都已经是0了，最后我们只需要检查最后一行是不是全部为0就可以检查这次的出操作是否正确了。如果正确且最小，那就存起来。最后输出，万事大吉。

当然，因为我们要改变第x行的1为0需要反转的是x+1行的位置。而这个整个规则是我们验证一组操作是否能达到目的所用的，那么我们需要在验证前先确定操作（没操作怎么验证..）。

于是根据规则内容可知，只需要能确认第一行的翻转情况，就能够推出下面所有的翻转情况并验证是否正确。于是需要做的就是枚举第一行的情况了。

【算法流程】

#include<stdio.h>
#include<string.h>
int n,m;
int count,min;
int s[20][20],s1[20][20];
int lg[20][20];
int cg(int a,int b)
{
int i,j;
s[a][b]=!s[a][b];
if(a-1>0)
s[a-1][b]=!s[a-1][b];
if(a+1<=n)
s[a+1][b]=!s[a+1][b];
if(b-1>0)
s[a][b-1]=!s[a][b-1];
if(b+1<=m)
s[a][b+1]=!s[a][b+1];
}
int do1(int a)
{
count=0;
int k=1;
int i,j;
memcpy(s,s1,sizeof(s));
memset(lg,0,sizeof(lg));
for(j=1;j<=m;j++)
{
if((a&(1<<(m-j)))>0) //记得a&1<<(m-j)要打括号，不然会是a&(1<<(m-j)>0)
{
cg(1,j),count++,lg[1][j]=1;
//printf("w%d %d\\n",a,a&(1<<(m-j)));
}
}
for(i=1;i<n;i++)
for(j=1;j<=m;j++)
{
if(s[i][j]==1)
cg(i+1,j),count++,lg[i+1][j]=1;
}
for(j=1;j<=m;j++)
if(s[n][j]==1)
k=0;
//if(k==1)
//printf("%d %d\\n",count,a);
return k;
}
int main()
{
int i,j,k;
while(scanf("%d%d",&n,&m)!=EOF)
{
min=n*m+1;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
scanf("%d",&s1[i][j]);
k=-1;
for(i=0;i<(1<<m);i++)//第一行翻动情况
{
if(do1(i)==1&&min>count)
{
min=count;
k=i;
//printf("%d\\n",min);
}
}
if(k==-1)
printf("IMPOSSIBLE\\n");
else
{
do1(k);
//printf("%d",k);
for(i=1;i<=n;i++)
{
j=1;
printf("%d",lg[i][j]);
for(j=2;j<=m;j++)
printf(" %d",lg[i][j]);
printf("\\n");
}
}
}
return 0;
}