Codeforces Round 656 (Div. 3)

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Codeforces Round 656 (Div. 3) EFG

E

题意：给有向边和无向边，确定无向边的方向使得没有环。

如果有向图有环，那么无解决

如果无环，显然可以拓扑排序，然后让无向边，指向拓扑序靠后的就永远不会有环。

代码 
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#include <bits/stdc++.h>
#include <set>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}
vector<pii> ans;
vector<pii> g[N];
int n, m;
int vis[N], in[N], v[N];
bool solve()
{
    queue<int> q;

    for (int i = 1; i <= n; i++)
    {
        if (!in[i])
            q.push(i);
    }

    while (!q.empty())
    {
        int x = q.front();
        q.pop();
        vis[x] = 1;
        for (pii now : g[x])
        {
            int to = now.first;
            if (now.second == 0)
            {
                in[to]--;
                ans.push_back(mk(x, to));
                if (!in[to])
                    q.push(to);
            }
            else if (!v[now.second])
            {
                v[now.second] = 1;
                if (!vis[to])
                    ans.push_back(mk(x, to));
                else
                    ans.push_back(mk(to, x));
            }
        }
    }
    for (int i = 1; i <= n; i++)
        if (!vis[i])
            return false;
    return true;
}

int main()
{
    int T = read();
    while (T--)
    {
        n = read(), m = read();
        for (int i = 1; i <= m; i++)
            v[i] = 0;
        for (int i = 1; i <= n; i++)
            vis[i] = 0, in[i] = 0, g[i].clear();
        ans.clear();
        for (int i = 1; i <= m; i++)
        {
            int t = read(), x = read(), y = read();
            if (t)
                g[x].push_back(mk(y, 0)), in[y]++;
            else
                g[x].push_back(mk(y, i)), g[y].push_back(mk(x, i));
        }
        if (!solve())
        {
            puts("NO");
        }
        else
        {
            puts("YES");
            for (pii now : ans)
            {
                printf("%d %d\n", now.first, now.second);
            }
        }
    }
}

F

题意：每次删除正好$k$个同根的点和边。

队列模拟一下，注意要为被删掉为$k$的倍数才能算一次操作。

代码 
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#include <bits/stdc++.h>
#include <set>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}

vector<int> g[N];
int d[N], p[N], vis[N];

int main()
{
    int T = read();
    while (T--)
    {
        int n = read(), k = read();
        for (int i = 1; i <= n; i++)
            g[i].clear(), d[i] = 0, p[i] = 0, vis[i] = 0;
        for (int i = 1; i < n; i++)
        {
            int u = read(), v = read();
            d[v]++;
            d[u]++;
            g[u].push_back(v);
            g[v].push_back(u);
        }
        queue<int> q;
        for (int i = 1; i <= n; i++)
            if (d[i] == 1)
                q.push(i);
        int ans = 0;
        while (!q.empty())
        {
            int x = q.front();
            q.pop();
            vis[x] = 1;
            for (int to : g[x])
            {
                if (vis[to] == 1)
                    continue;
                p[to]++;
                d[to]--;
                if (p[to] % k == 0)
                {
                    ans++;
                    if (d[to] == 1)
                        q.push(to);
                }
            }
        }
        printf("%d\n", ans);
    }
}

G

题意：两个数列，可以交换对应的坐标一次，就使得两个都是$1…n$的排序。

显然每个数出现2次即可，能保证完成。又可以找到，某些数字可以独立成环，枚举环$size$，和相应需要的翻转的次数$cnt$。显然假设再翻转一次，也可以。比较下$size-cnt,cnt$即可。

代码 
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#include <bits/stdc++.h>
#include <set>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}

vector<int> g[N];
int vis[N];

int a[2][N];
int main()
{
    int T = read();
    while (T--)
    {
        int n = read();
        for (int i = 1; i <= n; i++)
            g[i].clear(), vis[i] = 0;
        for (int i = 1; i <= n; i++)
            a[0][i] = read(), g[a[0][i]].push_back(i);
        for (int i = 1; i <= n; i++)
            a[1][i] = read(), g[a[1][i]].push_back(i);
        bool flag = 1;
        for (int i = 1; i <= n; i++)
            if (g[i].size() != 2)
                flag = 0;
        if (flag)
        {
            vector<int> ans;
            for (int i = 1; i <= n; i++)
                if (!vis[i])
                {
                    if (a[0][i] == a[1][i])
                        continue;

                    int f = 0;
                    int x = a[0][i];
                    int p = i;
                    vector<pii> tmp;

                    int num = 0;
                    while (!vis[p])
                    {
                        vis[p] = 1;

                        int q = (g[x][0] == p ? g[x][1] : g[x][0]);

                        if (a[f][q] == x)
                        {
                            x = a[f ^ 1][q];
                            tmp.push_back(mk(q, 1));

                            num++;
                        }
                        else
                        {
                            x = a[f][q];
                            tmp.push_back(mk(q, 0));
                        }
                        p = q;
                    }

                    for (pii now : tmp)
                    {
                        if (now.second == (num <= tmp.size() - num))
                            ans.push_back(now.first);
                    }
                }
            printf("%d\n", ans.size());
            for (int x : ans)
                printf("%d ", x);
            printf("\n");
        }
        else
            puts("-1");
    }
}