UVA 1564 - Widget Factory（高斯消元)

Posted 2020-10-20 zhchoutai

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，求每一个零件的制作时间。还要推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。注意每一个方程都是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;
}