hivalric Blossom

标签：ace   pac   main   元素   push   tor   for   signed   两个栈   

原题链接

• 题意：很多链，然后让每条链中各个元素都要挨在一起，然后给他们每个元素分配层数，层数相同的要以元素大小从左往右。
• 题解：就是两个栈模拟一下贪心即可。
• 代码：

#include <algorithm>
#include <cstring>
#include <cstdio>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int N = 1e5 + 9;
int id[N];
int ans[N];
int stk[N];
int Color[N];
int vis[N];
int l[N], r[N];
vector<int> G[N];
signed main() {
    int n, m;cin >> n >> m;
    int u, v;
    for (int i = 1; i <= m; i ++) {
        cin >> u >> v;
        G[u].push_back(v);
    }
    int nn = 0;
    for (int i = 1; i <= n; i ++) {
        if (!vis[i]) {
            nn++;
            vis[i] = 1;
            int u = i;
            l[u] = r[u] = 1;
            id[u] = nn;
            if (G[u].size()) {
                while (G[u].size() != 0) {
                    r[u] = 0;
                    u = G[u][0];
                    id[u] = nn;
                    vis[u] = 1;
                    r[u] = 1;
                }
            }
        }
    }
    int top =nn;
    for (int i = top, j = 1; i >= 0; i--,j++) {stk[i] = j;}
    for (int i = 1; i <= n; i ++) {
        if (l[i]) {
            //cout << i << "L\n";
            Color[id[i]] = stk[top--];
        }
        if (r[i]){
            top++;
            //cout << i << "R\n";
            stk[top] = Color[id[i]];
        }
        ans[i] = Color[id[i]];
    }
    for (int i = 1; i <= n; i ++) {
        cout << ans[i] << " ";
    }
}

hivalric Blossom

标签：ace   pac   main   元素   push   tor   for   signed   两个栈   

原文地址：https://www.cnblogs.com/Xiao-yan/p/14713383.html