poj1753
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题目描述简单!,以前用dfs写的,写的很挫!WA了不知道多少次!,想想,只有16个格子,每个格子有2中状态,所以枚举。用1——2^16次方枚举!用数字的位数来改变原图!
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
bool map[5][5]={};
int biao[5][5]={};
bool data[5][5]={};
int judge()
{
for(int i = 0; i < 4; i++)
for(int j = 0; j < 4; j++)
if(data[i][j] != data[0][0])
return 0;
return 1;
}
void change(int i,int j)
{
if(i - 1 >= 0) data[i-1][j] = !data[i-1][j];
if(i + 1 < 4) data[i+1][j] = !data[i+1][j];
if(j - 1 >= 0) data[i][j-1] = !data[i][j-1];
if(j + 1 < 4) data[i][j+1] = ! data[i][j+1];
data[i][j] = !data[i][j];
}
int dfs(int n)
{
for(int i = 0; i < 4; i++) for(int j = 0; j < 4; j++) data[i][j] = map[i][j];
int sum = 0;
for(int i = 3; i >= 0; i--)
for(int j = 3; j >= 0; j--)
if(n >= biao[i][j])
{
change(i,j);sum++;
n = n - biao[i][j];
if(judge())
return sum;
}
return -1;
}
int main()
{
int sum = 1;
for(int i = 0; i < 4; i++)
for(int j = 0; j < 4; j++)
{
biao[i][j] = sum;
sum = sum * 2;
}
char c;
for(int i = 0; i < 4; i++)
for(int j = 0; j < 4; j++)
{
scanf(" %c",&c);
if(c == 'w')
map[i][j] = 1;
else
map[i][j] = 0;
}
for(int i = 0; i < 4; i++) for(int j = 0; j < 4; j++) data[i][j] = map[i][j];
if(judge()) printf("0\n");
else{
int ans = sum;
for(int i = 1; i < sum; i++)
{
int t = dfs(i);
if(t == -1)
continue;
if(t < ans)
ans = t;
}
if(ans == sum)
printf("Impossible\n");
else
printf("%d\n",ans);}
return 0;
}
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