D. Gourmet choice（並查集 + 拓撲排序）

• 題目鏈接：http://codeforces.com/contest/1131/problem/D
• 題意：給你兩個數n,m分別表示兩組數的個數，第一組n個數，第二組m個數。給你兩組數之間的兩兩大小關係(>, = , <)，構造出一個滿足這些關係並且最大值最小的兩組數。
• 算法：並查集 + 拓撲排序

#include <bits/stdc++.h>
#define pi acos(-1.0 )
#define fastio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int INF = 0x3f3f3f3f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;//4e18 ~= 2^62
const int maxn =1000 + 10;
const LL mod = 1e9+7;

char s[maxn][maxn];
int par[maxn*2], ans[maxn*2], degree[maxn*2], vis[maxn*2];
vector<int> vec[maxn*2];
stack<int> stk;

int findr(int x)
{
    return par[x]==x? x:findr(par[x]);
}
int unite(int x, int y)
{
    x = findr(x);
    y = findr(y);
    if(x!=y) par[x] = y;
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i=1; i<=n; i++){
        scanf("%s", s[i]+1);
    }
    for(int i=1; i<=n+m; i++) par[i] = i;
    for(int i=1; i<=n; i++){
        for(int j=1; j<=m; j++){
            if(s[i][j] == '=') unite(i, j+n);
        }
    }
    for(int i=1; i<=n; i++){
        for(int j=1; j<=m; j++)
        if(s[i][j] == '>'){
            vec[findr(j+n)].push_back(findr(i)); degree[findr(i)]++;
        }
        else if(s[i][j] == '<') vec[findr(i)].push_back(findr(j+n)), degree[findr(j+n)]++;
    }

    for(int i=1; i<=n+m; i++){
        if(degree[findr(i)] == 0) {
            stk.push(findr(i)); vis[findr(i)] = 1;
            ans[findr(i)] = 1;
        }
    }
    int tmp;
    while(!stk.empty()){
        tmp = stk.top(); stk.pop();
        for(int v: vec[tmp]){
            degree[findr(v)]--;
            if(degree[findr(v)] == 0 && !vis[findr(v)]){
                stk.push(findr(v)); vis[findr(v)] = 1;
                ans[findr(v)] = ans[findr(tmp)]+1;
            }
        }
    }
    for(int i=1; i<=n+m; i++){
        if(!vis[findr(i)]) {
            printf("No\n");
            return 0;
        }
    }
    printf("Yes\n");
    for(int i=1; i<=n; i++) printf("%d%c", ans[findr(i)], " \n"[i==n]);
    for(int i=1; i<=m; i++) printf("%d%c", ans[findr(i+n)], " \n"[i==m]);
}