【高斯消元 && 模线性方程】

Step1 Problem:

题意表述很好直接用了
给出零件的种类数量n与记录的条数m，紧接着有m条记录，记录了在星期几到星期几之间（有可能间隔多个星期）成产了多少个什么样的零件。求每个零件生产需要多少天。
输入：
2 3
2 MON THU
1 2
3 MON FRI
1 1 2
3 MON SUN
1 2 2
输出：
8 3
数据范围：
1<=n, m<=300. 1 <= k <= 10000

Step2 Ideas:

(0*x1 + x2)%7 = (q[“THU”] - q[“MON”] + 1)
(1*x1 + x2)%7 = (q[“MON”] - q[“FRI”] + 1)
(0*x1 + 2*x2)%7 = (q[“SUN”] - q[“MON”] + 1)
对于上面样例 列出方程如上，高斯消元求解即可

Step3 Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<string>
#include<iostream>
using namespace std;
const int N = 350;
const int MOD = 7;
int data[N][N], lib[N], ans[N];
map<string, int> q;
int lcm(int x, int y)
{
    return x * y / __gcd(x, y);
}
int Gauss(int n)
{
    int rst = 0, r = 0;
    memset(lib, 0, sizeof(lib));
    for(int i = 0; i < n; i++)
    {
        for(int j = r; j < n; j++)
        {
            if(!data[j][i]) continue;
            for(int k = 0; k <= n; k++)
                swap(data[j][k], data[r][k]);
            break;
        }
        if(!data[r][i]) continue;
        for(int j = 0; j < n; j++)//利用r消去其他行，和01矩阵不同的地方
        {
            if(j == r || !data[j][i]) continue;
            int t = lcm(data[r][i], data[j][i]);
            int t1 = t/data[r][i], t2 = t/data[j][i];
            for(int k = 0; k <= n; k++)
                data[j][k] = ((data[j][k]*t2 - data[r][k]*t1)%MOD+MOD)%MOD;
        }
        lib[i] = 1; r++; rst++;
    }
    for(int i = 0; i < n; i++)
    {
        if(data[i][n]) {
            int flag = 0;
            for(int j = 0; j < n; j++)
            {
                if(data[i][j] != 0) {
                    flag = 1;
                    break;
                }
            }
            if(!flag) {
                return -1;
            }
        }
        if(!lib[i]) continue;
        for(int j = 0; j < n; j++)//如果唯一解 求解
        {
            if(data[j][i]) {
                int t = data[j][n];
                while(t % data[j][i]) t += MOD;
                ans[i] = t/data[j][i] % MOD;
            }
        }
    }
    return rst;
}
int main()
{
    q["MON"] = 1; q["TUE"] = 2;
    q["WED"] = 3; q["THU"] = 4;
    q["FRI"] = 5; q["SAT"] = 6; q["SUN"] = 7;
    int n, m, k, t;
    char st[15], et[15];
    while(~scanf("%d %d", &n, &m) && (n||m))
    {
        memset(data, 0, sizeof(data));
        int num = max(n, m);
        for(int i = 0; i < m; i++)
        {
            scanf("%d %s %s", &k, st, et);
            data[i][num] = ((q[et] - q[st] + 1)%MOD + MOD)%MOD;
            for(int j = 0; j < k; j++) {
                scanf("%d", &t);
                data[i][t-1]++;
                data[i][t-1] %= MOD;
            }
        }
        t = Gauss(num);
        if(t == -1) printf("Inconsistent data.\n");//无解
        else if(n > t) printf("Multiple solutions.\n");//多个解
        else {
            for(int i = 0; i < n; i++)
            {
                if(ans[i] < 3) ans[i] += MOD;
                if(i) printf(" ");
                printf("%d", ans[i]);
            }
            printf("\n");
        }
    }
    return 0;
}