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] = ;
     for (int s : v[i]) {
         if (used[s] == ) {
             puts("No");
             exit();
         }
         if (!used[s])    dfs(s);
     }
     used[i] = ;
     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\n", 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 = , s = , c = gc;
     while (c <= )    c = gc;
     if (c == '-')    s = -, c = gc;
     for (; isdigit(c); c = gc)
         x = x *  + c - ;
     return x * s;
 }

 const int maxn = 1e3 + ;
 int n, m;
 int value[maxn << ];
 int used[maxn << ];
 int fa[maxn << ];
 char cmp[maxn][maxn];
 vector<int> v[maxn << ], 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] = ;
     for (int s : v[i]) {
         if (used[s] == ) {
             puts("No");
             exit();
         }
         if (!used[s])    dfs(s);
     }
     used[i] = ;
     order.push_back(i);
 }

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

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

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

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

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