UVA 1564 - Widget Factory（高斯消元)

本文主要是介绍UVA 1564 - Widget Factory（高斯消元)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

UVA 1564 - Widget Factory

题目链接

题意：n种零件, 给定m个制作时间，每段时间制作k个零件，每种零件有一个制作时间，每段时间用Mon到Sun表示，求每个零件的制作时间，还要判断一下多解和无解的情况

思路：对于每段时间列出一个方程，这样一共列出m个方程解n个变元，利用高斯消元去求解，注意每个方程都是MOD 7的，所以在高斯消元过程中遇到除法要求该数字%7的逆元去进行运算

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;const int N = 305;
char week[7][5] = {"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"};int n, m, k, A[N][N], cnt[N];
char day1[5], day2[5];int find(char *day) {for (int i = 0; i < 7; i++)if (strcmp(week[i], day) == 0)return i;
}int inv(int x) {int ans = 1;for (int i = 0; i < 5; i++)ans = ans * x % 7;return ans;
}void build() {for (int i = 0; i < m; i++) {scanf("%d%s%s", &k, day1, day2);A[i][n] = (find(day2) - find(day1) + 8) % 7;int tmp;memset(cnt, 0, sizeof(cnt));while (k--) {scanf("%d", &tmp);cnt[tmp]++;}for (int j = 1; j <= n; j++)A[i][j - 1] = cnt[j] % 7;}
}int gauss() {int i = 0, j = 0;while (i < m && j < n) {int r;for (r = i; r < m; r++)if (A[r][j]) break;if (r == m) {j++;continue;}for (int k = j; k <= n; k++) swap(A[r][k], A[i][k]);for (int k = 0; k < m; k++) {if (i == k) continue;if (A[k][j]) {int tmp = A[k][j] * inv(A[i][j]) % 7;for (int x = j; x <= n; x++)A[k][x] = ((A[k][x] - tmp * A[i][x]) % 7 + 7) % 7;}}i++;}for (int k = i; k < m; k++)if (A[k][n]) return 2;if (i < n) return 1;for (int i = 0; i < n; i++) {int ans = A[i][n] * inv(A[i][i]) % 7;if (ans < 3) ans += 7;printf("%d%c", ans, i == n - 1 ? '\n' : ' ');}return 0;
}void solve() {int tmp = gauss();if (tmp == 1) printf("Multiple solutions.\n");else if (tmp == 2) printf("Inconsistent data.\n");
}int main() {while (~scanf("%d%d", &n, &m) && n) {build();solve();}return 0;
}

这篇关于UVA 1564 - Widget Factory（高斯消元)的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---