POJ1681 Painter's Problem (高斯消元or 状态dp)
Painter's Problem
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 4722 | Accepted: 2304 |
Description
There is a square wall which is made of n*n small square bricks. Some bricks are white while some bricks are yellow. Bob is a painter and he wants to paint all the bricks yellow. But there is something wrong with Bob's brush. Once he uses this brush to paint brick (i, j), the bricks at (i, j), (i-1, j), (i+1, j), (i, j-1) and (i, j+1) all change their color. Your task is to find the minimum number of bricks Bob should paint in order to make all the bricks yellow.
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Input
The first line contains a single integer t (1 <= t <= 20) that indicates the number of test cases. Then follow the t cases. Each test case begins with a line contains an integer n (1 <= n <= 15), representing the size of wall. The next n lines represent the original wall. Each line contains n characters. The j-th character of the i-th line figures out the color of brick at position (i, j). We use a 'w' to express a white brick while a 'y' to express a yellow brick.
Output
For each case, output a line contains the minimum number of bricks Bob should paint. If Bob can't paint all the bricks yellow, print 'inf'.
Sample Input
2 3 yyy yyy yyy 5 wwwww wwwww wwwww wwwww wwwww
Sample Output
0 15
思路：跟poj1225差不多..只不过这是给了初始状态,并且有多个解求出最小的那个,除了可以用高斯消元我用了状态dp写了下。
代码：(高斯消元)
View Code#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
//对2取模的01方程组
const int MAXN = 300;
//有equ个方程，var个变元。增广矩阵行数为equ,列数为var+1,分别为0到var
int equ,var;
int a[MAXN][MAXN]; //增广矩阵
int x[MAXN]; //解集
int free_x[MAXN];//用来存储自由变元（多解枚举自由变元可以使用）
int free_num;//自由变元的个数
//返回值为-1表示无解，为0是唯一解，否则返回自由变元个数
int Gauss()
{
int max_r,col,k;
free_num = 0;
for(k = 0, col = 0 ; k < equ && col < var ; k++, col++)
{
max_r = k;
for(int i = k+1; i < equ; i++)
{
if(abs(a[i][col]) > abs(a[max_r][col]))
max_r = i;
}
if(a[max_r][col] == 0)
{
k--;
free_x[free_num++] = col;//这个是自由变元
continue;
}
if(max_r != k)
{
for(int j = col; j < var+1; j++)
swap(a[k][j],a[max_r][j]);
}
for(int i = k+1; i < equ; i++)
{
if(a[i][col] != 0)
{
for(int j = col; j < var+1; j++)
a[i][j] ^= a[k][j];
}
}
}
for(int i = k; i < equ; i++)
if(a[i][col] != 0)
return -1;//无解
//if(k < var) return var-k;//自由变元个数
//唯一解，回代
for(int i = var-1; i >= 0; i--)
{
x[i] = a[i][var];
for(int j = i+1; j < var; j++)
x[i] ^= (a[i][j] && x[j]);
}
return 0;
}
int n;
void init()
{
memset(a,0,sizeof(a));
memset(x,0,sizeof(x));
equ = n*n;
var = n*n;
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
{
int t = i*n+j;
a[t][t] = 1;
if(i > 0)a[(i-1)*n+j][t] = 1;
if(i < n-1)a[(i+1)*n+j][t] = 1;
if(j > 0)a[i*n+j-1][t] = 1;
if(j < n-1)a[i*n+j+1][t] = 1;
}
}
char str[30][30];
int main()
{
//freopen("in.txt","r",stdin);
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
init();
for(int i = 0; i < n; i++)
{
scanf("%s",str[i]);
for(int j = 0; j < n; j++)
{
if(str[i][j] == 'y')
a[i*n+j][n*n] = 0;
else a[i*n+j][n*n] = 1;
}
}
int t = Gauss();
if(t == -1)
printf("inf\n");
else
{
int ans = 0;
for(int i = 0; i < n*n; i++)//因为x数组刚开始都清0了,这里没有设置自由未知量默认0代入(因为要最小)
ans += x[i];
printf("%d\n",ans);
}
}
return 0;
}
代码：(状压dp)
View Code#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int N = 20;
bool wall[N][N],pa[N][N];
int cou;
bool check(int n)
{
cou=0;
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
if(wall[i][j]^pa[i][j]^pa[i][j-1]^pa[i][j+1]^pa[i-1][j]^pa[i+1][j])return false;
if(pa[i][j])cou++;
}
}
return true;
}
int main()
{
//freopen("in.txt","r",stdin);
int t,n;
char s[N];
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(wall,0,sizeof(wall));
for(int i=1; i<=n; i++)
{
cin >> s;
for(int j=0; j<n; j++)
{
if(s[j]=='w')wall[i][j+1]=true;
}
}
int maxn = 1 << n, ans = 10000;
bool flag = false;
for(int i=0; i<maxn; i++)//枚举状态
{
memset(pa,0,sizeof(pa));
int tmp = i;
for(int j=1; j<=n; j++)
{
pa[1][j] = tmp & 1;
tmp = tmp >> 1;
}
for(int j=2; j<=n; j++)//第二行开始
for(int k=1; k<=n; k++)
{
if(wall[j-1][k] ^ pa[j-1][k] ^ pa[j-1][k-1] ^ pa[j-1][k+1] ^ pa[j-2][k])
//假如到pa[j-1][k]的左右上和本格和输入的结果为true,也就是还是白的,pa[j-1][k]的下要涂一次,也就是从第1行推第2行及以后
//异或0是不会变的..最后一行或边上会异或已经清零的多余格子..不影响
pa[j][k]=true;
}
if(check(n))
{
flag = true;
ans = min(ans, cou);
}
}
if(flag)
cout << ans << endl;
else
cout <<"inf" << endl;
}
return 0;
}
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