Codeforces #541 (Div2) - D. Gourmet choice（拓扑排序+并查集）

Problem   Codeforces #541 (Div2) - D. Gourmet choice

Time Limit: 2000 mSec

Problem Description

 

Input

Output

The first line of output should contain "Yes", if it's possible to do a correct evaluation for all the dishes, or "No" otherwise.

If case an answer exist, on the second line print nn integers — evaluations of dishes from the first set, and on the third line print mm integers — evaluations of dishes from the second set.

Sample Input

 3 4
>>>>
>>>>
>>>>

Sample Output

Yes
2 2 2
1 1 1 1

 

题解：一看就是拓扑排序，一通敲之后发现第三个样例过不去，原因是没有妥善处理等号的问题，其实很容易解决，相等的节点缩成一个，用并查集很容易实现，之后再拓扑排序就没问题了。

 

#include <bits/stdc++.h>

using namespace std;

#define REP(i, n) for (int i = 1; i <= (n); i++)
#define sqr(x) ((x) * (x))

const int maxn = 2000 + 10;
const int maxm = 200000 + 100;
const int maxs = 10000 + 10;

typedef long long LL;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;

const LL unit = 1LL;
const int INF = 0x3f3f3f3f;
const LL mod = 1000000007;
const double eps = 1e-14;
const double inf = 1e15;
const double pi = acos(-1.0);

int n, m;
string str[maxn];
vector<int> G[maxn];
int deg[maxn], ans[maxn];
int fa[maxn];
int Min, tot;

int findn(int x)
{
    return x == fa[x] ? x : fa[x] = findn(fa[x]);
}

void merge(int x, int y)
{
    int fx = findn(x), fy = findn(y);
    if (fx != fy)
    {
        fa[fx] = fy;
    }
}

bool toposort()
{
    queue<int> que;
    Min = 0;
    int cnt = 0;
    for (int i = 0; i < n + m; i++)
    {
        if (findn(i) == i && !deg[i])
        {
            que.push(i);
            cnt++;
            ans[i] = 0;
        }
    }
    while (!que.empty())
    {
        int u = que.front();
        que.pop();
        for (auto v : G[u])
        {
            int fv = findn(v);
            if (fv == v)
            {
                deg[fv]--;
                if (!deg[fv])
                {
                    cnt++;
                    que.push(fv);
                    ans[fv] = ans[u] - 1;
                    Min = min(ans[fv], Min);
                }
            }
        }
    }
    return cnt == tot;
}

void output()
{
    cout << "YES" << endl;
    for (int i = 0; i < n; i++)
    {
        cout << ans[findn(i)] - Min + 1;
        if (i != n - 1)
        {
            cout << " ";
        }
        else
        {
            cout << endl;
        }
    }
    for (int i = n; i < n + m; i++)
    {
        cout << ans[findn(i)] - Min + 1;
        if (i != n - 1)
        {
            cout << " ";
        }
        else
        {
            cout << endl;
        }
    }
}

void premanagement()
{
    for (int i = 0; i < n + m; i++)
    {
        if (findn(i) == i)
        {
            tot++;
        }
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    cin >> n >> m;
    for(int i = 0; i < n + m; i++)
    {
        fa[i] = i;
    }
    for (int i = 0; i < n; i++)
    {
        cin >> str[i];
    }
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (str[i][j] == '=')
            {
                merge(i, n + j);
            }
        }
    }
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (str[i][j] == '>')
            {
                int fi = findn(i), fj = findn(n + j);
                G[fi].push_back(fj);
                deg[fj]++;
            }
            else if (str[i][j] == '<')
            {
                int fi = findn(i), fj = findn(n + j);
                G[fj].push_back(fi);
                deg[fi]++;
            }
        }
    }
    premanagement();
    bool ok = toposort();
    if (!ok)
    {
        cout << "NO" << endl;
    }
    else
    {
        output();
    }
    return 0;
}