POJ 1753 高斯消元

題意是把所有的字母都變成w或者b的最小步數.

跑兩次01方程,取最小如果兩次都無解就輸出無解.板子題~

#include #include #include #include #include using namespace std;

#define maxn 333
#define free free
#define n 4
int a[maxn][maxn];
char mp[maxn][maxn];
int free[maxn*maxn]; //標記自由元
int x[maxn*maxn], cnt; //解集 自由元個數
int gauss () 
if (max_r != k) 
if (a[k][col] == 0) 
for (int i = k+1; i < n*n; i++) }}
}for (int i = k; i < n*n; i++) 
if (k < n*n)
return n*n - k; //返回自由元個數
for (int i = n*n-1; i >= 0; i--) 
return 0;
}int solve () 
else if (tot == 0) 
else 
num += x[id];
}ans= min (ans, num);
}return ans;
}}int main () 
memset (a, 0, sizeof a);
for (int i = 0; i < n; i++) 
}for (int i = 0; i < n; i++) 
}int ans1 = solve ();
memset (a, 0, sizeof a);
for (int i = 0; i < n; i++) 
}for (int i = 0; i < n; i++) 
}int ans2 = solve ();
if (ans1 == -1 && ans2 == -1)
cout << "impossible" << endl;
else if (ans1 == -1)
cout << ans2 << endl;
else if (ans2 == -1)
cout << ans1 << endl;
else cout << min (ans1, ans2) << endl;
}return 0;
}

高斯消元（poj1753

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