大致题意
把一个图分成三块，要求任意两块之间是完全图，块内部没有连线
分析
首先根据块内没有连线可以直接分成两块
假定点1是属于块1的，那么所有与点1连接的点，都不属于块1；反之则是块1的
然后在所有不属于块1的点内随意找一点k，设定其属于块2，那么所有与点k连接的点且不属于块1，则是块3。
块分完了，然后是判断每个块是否满足条件，我通过下面三条来判断
AC Code
（暴力就完事）
#include <bits/stdc++.h>
using namespace std;
#define MAXN 101000
int fa[MAXN]; // 保存了点属于哪个块
int deg[MAXN]; // 保存了点的度
pair<int, int> edge[MAXN * 3];
void solve() {
int n, m;
cin >> n >> m;
int f2 = 2; // f2 用来找块2
for (int i = 0; i < m; ++i) {
int u, v;
cin >> u >> v;
deg[u]++;
deg[v]++;
edge[i] = {u, v};
if (u == 1) {
fa[v] = 1;
f2 = v;
} else if (v == 1) {
fa[u] = 1;
f2 = u;
}
}
// 找出第三块
for (int i = 0; i < m; ++i) {
if (edge[i].first == f2 && fa[edge[i].second] == 1)
fa[edge[i].second] = 2;
else if (edge[i].second == f2 && fa[edge[i].first] == 1)
fa[edge[i].first] = 2;
}
int cnt[3] = {n, n, n}; // 保存了每个块内点的个数
// 需要变成完全图需要多少条边
for (int i = 0; i < n; ++i)
cnt[fa[i + 1]]--;
// 块内的入度是否符合条件
for (int i = 0; i < n; ++i) {
if (deg[i + 1] != cnt[fa[i + 1]]) {
cout << -1 << endl;
return;
}
}
// 每个块是否为空
if (cnt[0] == n || cnt[1] == n || cnt[2] == n) {
cout << -1 << endl;
return;
}
// 内部连线
for (int i = 0; i < m; ++i) {
if (fa[edge[i].first] == fa[edge[i].second]) {
cout << -1 << endl;
return;
}
}
for (int i = 0; i < n - 1; ++i)
cout << fa[i + 1] + 1 << " ";
cout << fa[n] + 1 << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
#ifdef ACM_LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
long long test_index_for_debug = 1;
char acm_local_for_debug;
while (cin >> acm_local_for_debug) {
cin.putback(acm_local_for_debug);
if (test_index_for_debug > 20) {
throw runtime_error("Check the stdin!!!");
}
auto start_clock_for_debug = clock();
solve();
auto end_clock_for_debug = clock();
cout << "Test " << test_index_for_debug << " successful" << endl;
cerr << "Test " << test_index_for_debug++ << " Run Time: "
<< double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
cout << "--------------------------------------------------" << endl;
}
#else
solve();
#endif
return 0;
}
总之一句话，暴力就完事了。反正边不多，我已经懒得优化了