Day 11

题目描述

题目很长，就不赘述了（主要是懒得写）

题目解析

Gauss 消元
题目的提示很明显，将元素守恒作为建立等式的基础。只要满足每一行元素守恒，即$x_1 + x_2 + ··· + x_n = 0$即可

元素个数为$m$，物质个数为$n$，增广矩阵的大下为$m * (n + 1)$，Gauss消元时间复杂度为$O(m n^2)$ 数据量很小

要注意的是，这里有个特解，比如$x_1 = x_2 = x_3 = ··· = x_n = 0$一定是成立的，但是在题干描述中并不合法，所以在$rankA = n$时还是要求出具体的解，判断一下特解

C++代码

#include <bits/stdc++.h>

using namespace std;
const int N = 110;
const double eps = 1e-10;

int T;
int n;
map<string , int> mp;
int idx = 0;
double g[N][N];

void process_str(int k , string x)
{
    int i = 0;
    while(i < x.size())
    {
        string s = "";
        while(i < x.size() && !isdigit(x[i])) s += x[i ++];
        int amount = 0;
        while(i < x.size() && isdigit(x[i])) amount = amount * 10 + x[i ++] - '0';

        if(!mp.count(s)) mp[s] = idx ++;

        g[mp[s]][k] = amount;
    }
}

int gauss()
{
    int c, r;
    for (c = 0, r = 0; c < n; c ++ )
    {
        int t = r;
        for (int i = r; i < idx; i ++ )
            if (fabs(g[i][c]) > fabs(g[t][c]))
                t = i;

        if (fabs(g[t][c]) < eps) continue;

        for (int i = c; i <= n; i ++ ) swap(g[t][i], g[r][i]);
        for (int i = n; i >= c; i -- ) g[r][i] /= g[r][c];
        for (int i = r + 1; i < idx; i ++ )
            if (fabs(g[i][c]) > eps)
                for (int j = n; j >= c; j -- )
                    g[i][j] -= g[r][j] * g[i][c];

        r ++ ;
    }

    if (r < n)
    {
        for (int i = r; i < idx; i ++ )
            if (fabs(g[i][n]) > eps)
                return 0; // 无解
        return 1;
    }

    for (int i = idx - 1; i >= 0; i -- )
        for (int j = i + 1; j < n; j ++ )
            g[i][n] -= g[i][j] * g[j][n];

    bool f = true;
    for(int i = 0 ; i < idx ; i ++)
        f &= (g[i][n] < eps);

    if(f) return 0;
    return 1;
}



void solve()
{
    memset(g , 0 , sizeof g);
    mp.clear();
    idx = 0;
    cin >> n;
    for(int i = 0 ; i < n ; i ++)
    {
        string x;
        cin >> x;
        process_str(i , x);    
    }

    if(gauss()) cout << "Y\n";
    else cout << "N\n";
}

int main()
{
    cin >> T;
    while(T --)
    {
        solve();
    }

    return 0;
}