高斯消元几道入门题总结POJ1222&&POJ1681&&POJ1830&&POJ2065&&POJ3185

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\n", 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("\n");}}}//system("pause");return 0;}

POJ1681：http://poj.org/problem?id=1681

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\n");}else{printf("%d\n",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));elsereturn 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~!!\n");}else{printf("%d\n", 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;elseval[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]);elseprintf(" %d", x[i]);}printf("\n");}//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;}