最近在搞高斯消元，反正这些题要么是我击败了它们，要么就是这些题把我给击败了。现在高斯消元专题部分还有很多题，先把几道很简单的入门题总结一下吧。
专题：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=29538#overview
专题地址来源于kuangbin大神，多谢kuangbin大神总结的高斯消元的模板，菜鸟在此谢过啊。
POJ1222：http://poj.org/problem?id=1222
题意是有5行六列30个开关，然后每拨动一个开关就会影响到其相邻的开关的状态。给了一个初始01的状态，问最终能否所有的值都为1。
一开始对这个高斯消元真的是相当地难以理解，要把这30个开关想象成是30个方程，第i个开关状态对应于第i个方程的右边的值。左边的是所有能够影响到的开关的值。我自己一开始这块真的很难理解，直到跑了几次程序才弄懂。然后就是高斯消元求解。其实高斯消元解方程本身的原理并不难，但就像图论题一样的，难在于建立方程，并且方程本身有各种幺蛾子，比方说我现在面对的POJ1487。。。跑题了。。。
代码：
#pragma warning(disable:4996)
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
int res[35];
int val[35][35];
int templat[35][35];
void Gauss()
{
int row, col, i, j, k;
for (row = 1, col = 1; col <= 30 && row <= 30; col++, row++)
{
k = row;
while (val[k][col] == 0 && k <= 30)
k++;
if (k != row)
{
for (i = 1; i <= 31; i++)
{
swap(val[k][i], val[row][i]);
}
}
for (i = row + 1; i <= 30; i++)
{
if (val[i][col])
{
for (j = col; j <= 31; j++)
{
val[i][j] = val[i][j] ^ val[row][j];
}
}
}
}
for (i = 30; i >= 1; i--)
{
res[i] = val[i][31];
for (j = 30; j > i; j--)
{
res[i] ^= (val[i][j] && res[j]);
}
}
}
int main()
{
//freopen("i.txt", "r", stdin);
//freopen("o.txt", "w", stdout);
int t, test, i, j, k;
memset(templat, 0, sizeof(templat));
for (i = 0; i < 5; i++)
{
for (j = 1; j <= 6; j++)
{
templat[i * 6 + j][i * 6 + j] = 1;
if (i - 1 >= 0)
{
templat[i * 6 + j][(i - 1) * 6 + j] = 1;
}
if (i + 1 < 5)
{
templat[i * 6 + j][(i + 1) * 6 + j] = 1;
}
if (j - 1 >= 1)
{
templat[i * 6 + j][i * 6 + j - 1] = 1;
}
if (j + 1 <= 6)
{
templat[i * 6 + j][i * 6 + j + 1] = 1;
}
}
}
scanf("%d", &t);
for (test = 1; test <= t; test++)
{
printf("PUZZLE #%d
", test);
memcpy(val, templat, sizeof(templat));
memset(res, 0, sizeof(res));
for (i = 0; i < 5; i++)
{
for (j = 1; j <= 6; j++)
{
scanf("%d", &val[i * 6 + j][31]);
}
}
Gauss();
for (i = 1; i <= 30; i++)
{
if (i % 6 == 1)
{
printf("%d", res[i]);
}
else
{
printf(" %d", res[i]);
}
if (i % 6 == 0)
{
printf("
");
}
}
}
//system("pause");
return 0;
}
和POJ1222一样的题意，理解POJ1222这个题也就好理解了，求出每一个点的值，查1的个数。
代码：
#pragma warning(disable:4996)
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
int dir[5][2] = {
{0,0},
{-1,0},
{1,0},
{0,-1},
{0,1}
};
int n;
int x[300];
char val_c[300][300];
int val[300][300];
int Gauss()
{
int row, col, i, j, k, ans;
ans = 0;
for (row = 1, col = 1; col <= n&&row <= n; col++, row++)
{
k = row;
while (val[k][col] == 0 && k <= n)
k++;
if (k>n)
{
row--;
ans++;
continue;
}
if (k != row)
{
for (j = 1; j <= n + 1; j++)
{
swap(val[k][j], val[row][j]);
}
}
for (i = row + 1; i <= n; i++)
{
if (val[i][col])
{
for (j = col; j <= n + 1; j++)
{
val[i][j] = val[i][j] ^ val[row][j];
}
}
}
}
for (i = row; i <= n; i++)
{
if (val[i][n + 1])
return -1;
}
for (i = n - ans; i >= 1; i--)
{
x[i] = val[i][n + 1];
for (j = n; j > i; j--)
{
x[i] ^= (val[i][j] && x[j]);
}
}
ans = 0;
for (i = 1; i <= n; i++)
{
if (x[i])
ans++;
}
return ans;
}
int main()
{
//freopen("i.txt", "r", stdin);
//freopen("o.txt", "w", stdout);
int a, b, ans;
int t, i, j, k;
scanf("%d", &t);
while (t--)
{
scanf("%d", &n);
memset(val, 0, sizeof(val));
memset(val_c, 0, sizeof(val_c));
memset(x, 0, sizeof(x));
for (i = 1; i <= n; i++)
{
for (j = 1; j <= n; j++)
{
for (k = 0; k <= 4; k++)
{
a = i + dir[k][0];
b = j + dir[k][1];
if (a >= 1 && a <= n&&b >= 1 && b <= n)
{
val[(i - 1)*n + j][(a - 1)*n + b] = 1;
}
}
}
}
for (i = 1; i <= n; i++)
{
cin >> val_c[i] + 1;
for (j = 1; j <= n; j++)
{
if (val_c[i][j] == 'w')
{
val[(i - 1)*n + j][n*n + 1] = 1;
}
}
}
n = n*n;
ans = Gauss();
if (ans == -1)
{
printf("inf
");
}
else
{
printf("%d
",ans);
}
}
//system("pause");
return 0;
}POJ1830：http://poj.org/problem?id=1830
判断多解、无解的情况。另外这个题目一定要注意i、j的顺序，如果j的开关影响到了i，那么应该是在i的方程里面出现j，所以是val[i][j]=1。放到题目当中，两者顺序要调换！！！
代码：
#pragma warning(disable:4996)
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
int n;
int s[350], e[350], x[350], x2[350];
int res[350][350], res2[350][350];
int Gauss()
{
int row, col, i, j, k, ans;
ans = 0;
for (row = 1, col = 1; col <= n&&row <= n; col++, row++)
{
k = row;
while (res[k][col] == 0 && k <= n)
k++;
if (k>n)
{
row--;
ans++;
continue;
}
if (k != row)
{
for (j = 1; j <= n + 1; j++)
{
swap(res[k][j], res[row][j]);
}
}
for (i = row + 1; i <= n; i++)
{
if (res[i][col])
{
for (j = col; j <= n + 1; j++)
{
res[i][j] = res[i][j] ^ res[row][j];
}
}
}
}
for (i = row; i <= n; i++)
{
if (res[i][n + 1])
return -1;
}
if (row <= n)
return (1 << (n - row+1));
else
return 1;
}
int main()
{
//freopen("i.txt", "r", stdin);
//freopen("o.txt", "w", stdout);
int k, i, j, ans;
scanf("%d", &k);
while (k--)
{
scanf("%d", &n);
memset(res, 0, sizeof(res));
memset(res2, 0, sizeof(res2));
memset(x, 0, sizeof(x));
memset(x2, 0, sizeof(x2));
memset(e, 0, sizeof(e));
memset(s, 0, sizeof(s));
for (i = 1; i <= n; i++)
{
scanf("%d", &s[i]);
res[i][i] = 1;
}
for (i = 1; i <= n; i++)
{
scanf("%d", &e[i]);
res[i][n + 1] = s[i] ^ e[i];
}
for (;;)
{
scanf("%d%d", &i, &j);
if (i == 0 && j == 0)
break;
res[j][i] = 1;
}
memcpy(res2, res, sizeof(res));
ans = Gauss();
if (ans == -1)
{
printf("Oh,it's impossible~!!
");
}
else
{
printf("%d
", ans);
}
}
//system("pause");
return 0;
}POJ2065：http://poj.org/problem?id=2065
题意实在是。。。直接理解成来了一个质数p和一个字符串，这个字符串的长度决定了有多少个方程，每个方程的右边的值是相应的字符的值，a对应0，b对应1。。。z对应26，*对应0。
第i个方程左边第j个变量的系数是pow(i,j)mod p，直接求解。
代码：
#pragma warning(disable:4996)
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
int p;
char res[75];
int x[305];//解集
int val[305][305];//增广矩阵
bool free_x[305];//标记是否是不确定的变元
inline int gcd(int a, int b)
{
int t;
while (b != 0)
{
t = b;
b = a%b;
a = t;
}
return a;
}
inline int lcm(int a, int b)
{
return a / gcd(a, b)*b;//先除后乘防溢出
}
int Gauss(int equ, int var)
{
int i, j, k;
int max_r;//当前这列绝对值最大的行
int col;//当前处理的列
int ta, tb;
int LCM;
int temp;
int free_x_num;
int free_index;
for (int i = 0; i <= var; i++)
{
x[i] = 0;
free_x[i] = true;
}
//转换为阶梯阵
col = 0;//当前处理的列
for (k = 0; k < equ&&col < var; k++, col++)
{
//枚举当前处理的行
//找到该col列元素绝对值最大的那行与第k行交换.(为了在除法时减少误差)
max_r = k;
for (i = k + 1; i < equ; i++)
{
if (abs(val[i][col])>abs(val[max_r][col]))
max_r = i;
}
if (max_r != k)
{//与第k行交换
for (j = k; j < var + 1; j++)
swap(val[k][j], val[max_r][j]);
}
if (val[k][col] == 0)
{
k--;
continue;
}
for (i = k + 1; i < equ; i++)
{//枚举要删去的行
if (val[i][col] != 0)
{
LCM = lcm(abs(val[i][col]), abs(val[k][col]));
ta = LCM / abs(val[i][col]);
tb = LCM / abs(val[k][col]);
if (val[i][col] * val[k][col] < 0)
tb = -tb;
for (j = col; j < var + 1; j++)
{
val[i][j] = ((val[i][j] * ta - val[k][j] * tb) % p + p) % p;
}
}
}
}
for (i = var - 1; i >= 0; i--)
{
temp = val[i][var];
for (j = i + 1; j < var; j++)
{
if (val[i][j] != 0)
{
temp = temp - val[i][j] * x[j];
temp = (temp % p + p) % p;
}
}
while ((temp % val[i][i]))temp += p;
x[i] = ((temp / val[i][i]) % p + p) % p;
}
return 0;
}
int getresult(int A, int n, int k)
{
int b = 1;
while (n > 0)
{
if (n & 1)
{
b = (b*A) % k;
}
n = n >> 1;
A = (A*A) % k;
}
return b;
}
int main()
{
//freopen("i.txt", "r", stdin);
//freopen("o.txt", "w", stdout);
int test, i, j, len;
scanf("%d", &test);
while (test--)
{
memset(val, 0, sizeof(val));
scanf("%d%s", &p, res);
len = strlen(res);
for (i = 0; i < len; i++)
{
if (res[i] == '*')
val[i][len] = 0;
else
val[i][len] = res[i] - 'a' + 1;
for (j = 0; j < len; j++)
val[i][j] = getresult(i + 1, j, p);
}
Gauss(len, len);
for (i = 0; i < len; i++)
{
if (i == 0)
printf("%d", x[i]);
else
printf(" %d", x[i]);
}
printf("
");
}
//system("pause");
return 0;
}POJ3185：http://poj.org/problem?id=3185
之前的好歹是个二维的，这次的这个是一维的，每个bowl翻过来影响到周围的两个，没有用高斯消元。。。太麻烦了，直接贪心。
代码：
#pragma warning(disable:4996)
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
int minn;
int val[25];
int val_c[25];
void dfs1(int i, int a[],int num)
{
if (i == 22)
{
minn = min(minn, num);
return;
}
if (a[i - 1] == 1)
{
a[i - 1] = (a[i - 1] + 1) & 1;
a[i] = (a[i] + 1) & 1;
a[i + 1] = (a[i + 1] + 1) & 1;
dfs1(i + 1, a, num + 1);
a[i - 1] = (a[i - 1] + 1) & 1;
a[i] = (a[i] + 1) & 1;
a[i + 1] = (a[i + 1] + 1) & 1;
}
else
{
dfs1(i + 1, a, num);
}
}
void dfs2(int i, int a[], int num)
{
if (i == -1)
{
minn = min(minn, num);
return;
}
if (a[i + 1] == 1)
{
a[i + 1] = (a[i + 1] + 1) & 1;
a[i] = (a[i] + 1) & 1;
a[i - 1] = (a[i - 1] + 1) & 1;
dfs2(i - 1, a, num + 1);
a[i + 1] = (a[i + 1] + 1) & 1;
a[i] = (a[i] + 1) & 1;
a[i - 1] = (a[i - 1] + 1) & 1;
}
else
{
dfs2(i - 1, a, num);
}
}
int main()
{
//freopen("i.txt", "r", stdin);
//freopen("o.txt", "w", stdout);
int i;
memset(val, 0, sizeof(val));
for (i = 1; i <= 20; i++)
scanf("%d", val + i);
minn = 200;
dfs1(2,val,0);
dfs2(19, val, 0);
cout << minn << endl;
//system("pause");
return 0;
}