codeforces 1228D Complete Tripartite (graph theory + thinking)

Portal

Meaning of the questions: Given a simple graph Unicom is not guaranteed, and asked whether there is staining protocols so that each communication block diagram for the three points.

Analysis: for no other constraints, considering only the third set inside each sub FIG unbound edges connected, a first dyed FIG. For the first approach is a communication block, to select a starting point u, the dyed u 1, u is then connected directly to the point, all dyed 2, and then select a dyed point 2 is v, then all v w staining point directly connected, wherein if w is not stained, then dyed w 1, w if 2 was dyed, the dyed w 3. And then determine the eligibility stained map.

For all Unicom blocks, the number of three kinds of patches to maintain the midpoint of the communication block, a link block number if there is a certain kind of color is 0, indicating that the three subsets Minato dissatisfaction communication block, not legal.

For a point, if the number of points he is directly connected to the current number of China Unicom block points and the other two colors, the legal, otherwise illegal.

For one side, if Unicom same color point, not legal.

Code determination may be redundant.

Code:

#include <bits/stdc++.h>
#define mp make_pair
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
typedef long long LL;
const int maxn = 3e5 + 10;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
using namespace std;
int n, m, u, v, col[maxn], f[maxn], sz[maxn];
int find(int x) {
    return f[x] == -1 ? x : f[x] = find(f[x]);
}
vector<int> g[maxn];
vector<int> s[maxn][3];
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 0; i < m; ++i) {
        cin >> u >> v;
        g[u].pb(v);
        g[v].pb(u);
    }
    for (int i = 1; i <= n; ++i)
        f[i] = -1;
    for (int i = 1; i <= n; ++i) {
        if (col[i] == 0) {
            col[i] = 1;
            int fu = find(i);
            if (g[i].empty()) {
                cout << -1;
                exit(0);
            }
            int to = g[i][0];
            for (int j = 0; j < g[i].size(); ++j) {
                int _to = g[i][j];
                int fv = find(_to);
                if (fv != fu)
                    f[fv] = fu;
                col[_to] = 2;
            }
            for (int j = 0; j < g[to].size(); ++j) {
                int _to = g[to][j];
                if (_to == i)
                    continue;
                int fv = find(_to);
                if (fv != fu)
                    f[fv] = fu;
                if (col[_to] == 0)
                    col[_to] = 1;
                else if (col[_to] == 2)
                    col[_to] = 3;
                else {
                    cout << -1;
                    exit(0);
                }
            }
        }
    }
    for (int i = 1; i <= n; ++i) {
        v = find(i);
        s[v][col[i] - 1].pb(i);
    }
    for (int i = 1; i <= n; ++i)
        if (!s[i][0].empty() || !s[i][1].empty() || !s[i][2].empty()) {
            if (s[i][0].empty() || s[i][1].empty() || s[i][2].empty()) {
                cout << -1;
                exit(0);
            }
        }
    for (int i = 1; i <= n; ++i) {
        v = find(i);
        if (col[i] == 1) {
            if (s[v][1].size() + s[v][2].size() != g[i].size()) {
                cout << -1;
                exit(0);
            }
        }
        else if (col[i] == 2) {
            if (s[v][0].size() + s[v][2].size() != g[i].size()) {
                cout << -1;
                exit(0);
            }
        }
        else if (col[i] == 3) {
            if (s[v][0].size() + s[v][1].size() != g[i].size()) {
                cout << -1;
                exit(0);
            }
        }
        for (int j = 0; j < g[i].size(); ++j) {
            int to = g[i][j];
            if (col[to] == col[i]) {
                cout << -1;
                exit(0);
            }
        }
    }
    for (int i = 1; i <= n; ++i)
        cout << col[i] << " ";
}

View Code