CF #541div2 D

题目本质：形成一个拓扑图，不应带自环。

解决方法：

1.先把等于号的部分用dsu缩点；

2.大于和小于号建立拓扑关系；

3.n*m的矩阵，只要用标号n+j代表m集合的第j个就从二维降到一维了；

4.dfs查有没有环：used == 2的那种环是合法的！

void dfs(int i) {
    used[i] = 1;
    for (int s : v[i]) {
        if (used[s] == 1) {
            puts("No");
            exit(0);
        }
        if (!used[s])    dfs(s);
    }
    used[i] = 2;
    order.push_back(i);
}

5.按照order记录的拓扑顺序自底向上dp一下最小取值。

总代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <cctype>
#include <climits>
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <string>
#include <sstream>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <list>
#include <fstream>
#define ri readint()
#define gc getchar()
#define R(x) scanf("%d", &x)
#define W(x) printf("%d
", x)
#define init(a, b) memset(a, b, sizeof(a))
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define irep(i, a, b) for (int i = a; i >= b; i--)
#define ls p << 1
#define rs p << 1 | 1
using namespace std;

typedef double db;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
const int inf = 0x3f3f3f3f;
const ll INF = 1e18;

inline int readint() {
    int x = 0, s = 1, c = gc;
    while (c <= 32)    c = gc;
    if (c == '-')    s = -1, c = gc;
    for (; isdigit(c); c = gc)
        x = x * 10 + c - 48;
    return x * s;
}

const int maxn = 1e3 + 5;
int n, m;
int value[maxn << 1];
int used[maxn << 1];
int fa[maxn << 1];
char cmp[maxn][maxn];
vector<int> v[maxn << 1], order;

int getf(int v) {
    return v == fa[v] ? v : fa[v] = getf(fa[v]);
}

void merge(int x, int y) {
    int p = getf(x), t = getf(y);
    fa[p] = t;
}

void dfs(int i) {
    used[i] = 1;
    for (int s : v[i]) {
        if (used[s] == 1) {
            puts("No");
            exit(0);
        }
        if (!used[s])    dfs(s);
    }
    used[i] = 2;
    order.push_back(i);
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    cin >> n >> m;
    rep(i, 1, n + m) {
        fa[i] = i;
    }

    rep(i, 1, n) {
        rep(j, 1, m) {
            cin >> cmp[i][j];
            if (cmp[i][j] == '=') {
                merge(i, n + j);
            }
        }
    }

    rep(i, 1, n) {
        rep(j, 1, m) {
            int x = getf(i), y = getf(n + j);
            if (cmp[i][j] == '>') {
                v[x].push_back(y);
            } else if (cmp[i][j] == '<') {
                v[y].push_back(x);
            }
        }
    }

    rep(i, 1, n + m) {
        if (fa[i] != i)
            continue;
        if (!used[i]) {
            dfs(i);
        }
    }
    for (int i : order) {
        int val = 0;
        for (int j : v[i]) {
            val = max(val, value[j]);
        }
        value[i] = val + 1;
    }

    cout << "Yes
";
    rep(i, 1, n) {
        cout << value[getf(i)] << " ";
    }
    cout << endl;
    rep(i, 1, m) {
        cout << value[getf(n + i)] << " ";
    }
    return 0;
}