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！

 1 void dfs(int i) {
 2     used[i] = 1;
 3     for (int s : v[i]) {
 4         if (used[s] == 1) {
 5             puts("No");
 6             exit(0);
 7         }
 8         if (!used[s])    dfs(s);
 9     }
10     used[i] = 2;
11     order.push_back(i);
12 }

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：

  1 #pragma comment(linker, "/STACK:1024000000,1024000000")
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <cctype>
  8 #include <climits>
  9 #include <iostream>
 10 #include <iomanip>
 11 #include <algorithm>
 12 #include <string>
 13 #include <sstream>
 14 #include <stack>
 15 #include <queue>
 16 #include <set>
 17 #include <map>
 18 #include <vector>
 19 #include <list>
 20 #include <fstream>
 21 #define ri readint()
 22 #define gc getchar()
 23 #define R(x) scanf("%d", &x)
 24 #define W(x) printf("%d
", x)
 25 #define init(a, b) memset(a, b, sizeof(a))
 26 #define rep(i, a, b) for (int i = a; i <= b; i++)
 27 #define irep(i, a, b) for (int i = a; i >= b; i--)
 28 #define ls p << 1
 29 #define rs p << 1 | 1
 30 using namespace std;
 31 
 32 typedef double db;
 33 typedef long long ll;
 34 typedef unsigned long long ull;
 35 typedef pair<int, int> P;
 36 const int inf = 0x3f3f3f3f;
 37 const ll INF = 1e18;
 38 
 39 inline int readint() {
 40     int x = 0, s = 1, c = gc;
 41     while (c <= 32)    c = gc;
 42     if (c == '-')    s = -1, c = gc;
 43     for (; isdigit(c); c = gc)
 44         x = x * 10 + c - 48;
 45     return x * s;
 46 }
 47 
 48 const int maxn = 1e3 + 5;
 49 int n, m;
 50 int value[maxn << 1];
 51 int used[maxn << 1];
 52 int fa[maxn << 1];
 53 char cmp[maxn][maxn];
 54 vector<int> v[maxn << 1], order;
 55 
 56 int getf(int v) {
 57     return v == fa[v] ? v : fa[v] = getf(fa[v]);
 58 }
 59 
 60 void merge(int x, int y) {
 61     int p = getf(x), t = getf(y);
 62     fa[p] = t;
 63 }
 64 
 65 void dfs(int i) {
 66     used[i] = 1;
 67     for (int s : v[i]) {
 68         if (used[s] == 1) {
 69             puts("No");
 70             exit(0);
 71         }
 72         if (!used[s])    dfs(s);
 73     }
 74     used[i] = 2;
 75     order.push_back(i);
 76 }
 77 
 78 int main() {
 79     ios_base::sync_with_stdio(false);
 80     cin.tie(0);
 81 
 82     cin >> n >> m;
 83     rep(i, 1, n + m) {
 84         fa[i] = i;
 85     }
 86 
 87     rep(i, 1, n) {
 88         rep(j, 1, m) {
 89             cin >> cmp[i][j];
 90             if (cmp[i][j] == '=') {
 91                 merge(i, n + j);
 92             }
 93         }
 94     }
 95 
 96     rep(i, 1, n) {
 97         rep(j, 1, m) {
 98             int x = getf(i), y = getf(n + j);
 99             if (cmp[i][j] == '>') {
100                 v[x].push_back(y);
101             } else if (cmp[i][j] == '<') {
102                 v[y].push_back(x);
103             }
104         }
105     }
106 
107     rep(i, 1, n + m) {
108         if (fa[i] != i)
109             continue;
110         if (!used[i]) {
111             dfs(i);
112         }
113     }
114     for (int i : order) {
115         int val = 0;
116         for (int j : v[i]) {
117             val = max(val, value[j]);
118         }
119         value[i] = val + 1;
120     }
121 
122     cout << "Yes
";
123     rep(i, 1, n) {
124         cout << value[getf(i)] << " ";
125     }
126     cout << endl;
127     rep(i, 1, m) {
128         cout << value[getf(n + i)] << " ";
129     }
130     return 0;
131 }