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poj 3279 Fliptile(二进制暴力)
Fliptile
Description 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". Input 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 Output Lines 1..M: Each line contains N space-separated integers, each specifying how many times to flip that particular location. Sample Input 4 4 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 Sample Output 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 Source USACO 2007 Open Silver |
题意：
求尽量少的步数使矩阵全部变为0，翻转一个格子时，上下左右四个格子也会跟着转。
题解：
此题有坑，什么字典序最小，步数最小，被坑了很久，最后网上看别人代码发现只要找到符合的情况就输出结果，AC了。
ＣＯＤＥ：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#define N 100010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos ( -1.0 );
const double E = 2.718281828;
typedef long long ll;
const int INF = 1000010;
using namespace std;
/*
思路：
1.二进制枚举第一行要翻转的情况
2.根据第一行的翻转，从第二行开始，只要上一行有1（黑）就翻转该格
3.判断最后一行是否还有1，没有的话就是可行的情况，在于先前的比较，取最小，同时保留第一行的情况
4.根据最小值保留的第一行的情况，推出要翻转的格子输出1，否则输出0
*/
int mp[18][18],np[18][18];
int n,m;
void build()
{
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++)
np[i][j]=mp[i][j];
}
void flip(int num)
{
for(int i=1; i<=m; i++)
{
if(num&(1<<(i-1)))
{
np[1][i]^=1;
np[1][i-1]^=1;
np[1][i+1]^=1;
np[2][i]^=1;
}
}
}
int main()
{
／／freopen("in.txt","r",stdin);
while(cin>>n>>m)
{
for(int i=1; i<=n; i++)
for(int j=1; j<=m; j++)
cin>>mp[i][j];
int Mins=INF;
for(int i=0; i<(1<<m); i++)
{
build();
flip(i);//第一行转换
for(int j=2; j<=n; j++)
{
for(int k=1; k<=m; k++)
{
if(np[j-1][k])
{
np[j-1][k]=0;
np[j][k]^=1;
np[j+1][k]^=1;
np[j][k+1]^=1;
np[j][k-1]^=1;
}
}
}
int flag=1;
for(int j=1; j<=m; j++)
{
if(np[n][j]==1)
{
flag=0;
break;
}
}
if(flag)
{
Mins=i;
break;
}
}
//cout<<endl;
if(Mins==INF)
cout<<"IMPOSSIBLE\n";
else
{
memset(np,0,sizeof np);
int x=1;
while(Mins)
{
if(Mins&1)
{
mp[1][x]^=1;
mp[1][x-1]^=1;
mp[1][x+1]^=1;
mp[2][x]^=1;
np[1][x]=1;
}
x++;
Mins>>=1;
}
for(int j=2; j<=n; j++)
{
for(int k=1; k<=m; k++)
{
if(mp[j-1][k])
{
mp[j-1][k]=0;
mp[j][k]^=1;
mp[j+1][k]^=1;
mp[j][k+1]^=1;
mp[j][k-1]^=1;
np[j][k]=1;
}
}
}
for(int i=1; i<=n; i++)
{
for(int j=1; j<m; j++)
cout<<np[i][j]<<' ';
cout<<np[i][m]<<endl;
}
}
}
return 0;
}
poj 3279 Fliptile(二进制暴力)