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2020-2021 ACM-ICPC Brazil Subregional Programming Contest K Between Us

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题目描述

https://codeforces.com/gym/102861/problem/K

Solution

我们令 $x_i=0/1$ 表示 $i$ 在哪个集合

如果 $i$ 的度数为奇数，则 $x_i~xor~x_a ~xor~ x_b~xor~ x_c\cdots=1$

如果 $i$ 的度数为偶数，则 $x_a ~xor~ x_b~xor~ x_c\cdots=0$

那么相当于我们求这个异或方程组是否有解，高消即可

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#include <iostream>
#include <cstdio>
#include <vector>
#define maxn 110
using namespace std;

int n, m, du[maxn];

vector<int> G[maxn];

int g[maxn][maxn];
void Gauss() {
    for (int i = 1; i <= n; ++i) {
        int p = -1;
        for (int j = i; j <= n; ++j) if (g[j][i]) p = j;
        if (p == -1) continue; swap(g[p], g[i]); 
        for (int j = 1; j <= n; ++j) {
            if (i == j || !g[j][i]) continue;
            for (int k = i; k <= n + 1; ++k) g[j][k] ^= g[i][k];
        }
    } 
    for (int i = 1; i <= n; ++i) {
        int s = 0; 
        for (int j = i; j <= n; ++j) s |= g[i][j];
        if (!s && g[i][n + 1]) return (void) (cout << "N\n");
    }
    cout << "Y\n";
} 

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n >> m;
    for (int i = 1; i <= m; ++i) {
        int x, y; cin >> x >> y;
        ++du[x]; ++du[y];
        G[x].push_back(y); G[y].push_back(x); 
    }
    for (int i = 1; i <= n; ++i) 
        if (du[i] % 2 == 0) {
            for (auto u : G[i]) g[i][u] = 1;
            g[i][n + 1] = 1; 
        }
        else {
            for (auto u : G[i]) g[i][u] = 1;
            g[i][i] = 1; g[i][n + 1] = 0;
        }
    Gauss();
    return 0;
}