2015年四川省赛||弱校联萌十一大决战之强力热身 H. Range Query 二分图的最大匹配确定匹配顺序

http://www.bnuoj.com/v3/contest_show.php?cid=6865#problem/H

frog has a permutation \(p(1), p(2), \dots, p(n)\) of \(\{1, 2, \dots, n\}\). She also has \(m_1 + m_2\) records \((a_i, b_i, c_i)\) of the permutation.

 

• For \(1 \leq i \leq m_1\), \((a_i, b_i, c_i)\) means \(\min\{p(a_i), p(a_i + 1), \dots, p(b_i)\} = c_i\);
• For \(m_1 < i \leq m_1 + m_2\), \((a_i, b_i, c_i)\) means \(\max\{p(a_i), p(a_i + 1), \dots, p(b_i)\} = c_i\).

Find a permutation which is consistent with above records, or report the records are self-contradictory. If there are more than one valid permutations, find the lexicographically least one.

 

Permutation \(p(1), p(2), \dots, p(n)\) is lexicographically smaller than \(q(1), q(2), \dots, q(n)\) if and only if there exists \(1 \leq i \leq n\) which \(p(i) < q(i)\) and for all \(1 \leq j < i\), \(p(j) = q(j)\).

Input

The input consists of multiple tests. For each test:

 

The first line contains \(3\) integers \(n, m_1, m_2\) (\(1 \leq n \leq 50, 0 \leq m_1 + m_2 \leq 50\)). Each of the following \((m_1 + m_2)\) lines contains \(3\) integers \(a_i, b_i, c_i\) (\(1 \leq a_i \leq b_i \leq n, 1 \leq c_i \leq n\)).

Output

For each test, write \(n\) integers \(p(1), p(2), \dots, p(n)\) which denote the lexicographically least permutation, or $``\texttt{-1}''$ if records are self-contradictory.

Sample Input

5 1 1
1 5 1
1 5 5
3 1 1
1 2 2
1 2 2

Sample Output

1 2 3 4 5
-1

/**
2015年四川省赛||弱校联萌十一大决战之强力热身 H. Range Query  二分图的最大匹配确定匹配顺序
题目大意：给定一个序列1~n，然后1~n个位置，现在给定一些子区间的最大最小值，问是否可以一个位置对应一个数，输出字典序最小的
解题思路：看叉姐的思路http://talk.icpc-camp.org/d/117-2015-h-range-query
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=155;
int n,m1,m2,low[maxn],up[maxn],start[maxn],tail[maxn];
int a[maxn][maxn];
int used[maxn],link[maxn],vis[maxn];
bool dfs(int u)
{
    for(int v=1;v<=n;v++)
    {
        if(a[u][v]&&!vis[v]&&!used[v])
        {
            vis[v]=1;
            if(link[v]==-1||dfs(link[v]))
            {
                link[v]=u;
                return true;
            }
        }
    }
    return false;
}
bool solve(int s)
{
    memset(link,-1,sizeof(link));
    for(int u=s;u<=n;u++)
    {
        memset(vis,0,sizeof(vis));
        if(!dfs(u))return false;
    }
    return true;
}
int main()
{
    while(~scanf("%d%d%d",&n,&m1,&m2))
    {
        for(int i=1; i<=n; i++)
        {
            low[i]=start[i]=1;
            up[i]=tail[i]=n;
        }
        for(int i=0; i<m1; i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            for(int j=a; j<=b; j++) low[j]=max(low[j],c);
            start[c]=max(start[c],a);
            tail[c]=min(tail[c],b);
        }
        for(int i=0;i<m2;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            for(int j=a;j<=b;j++)up[j]=min(up[j],c);
            start[c]=max(start[c],a);
            tail[c]=min(tail[c],b);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                a[i][j]=low[i]<=j&&j<=up[i]&&start[j]<=i&&i<=tail[j];
            }
        }
        memset(used,0,sizeof(used));
        if(solve(1))
        {
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=n;j++)
                {
                    if(a[i][j]&&!used[j])
                    {
                        used[j]=1;
                        if(solve(i+1))
                        {
                            printf("%d%c",j," \n"[i==n]);
                            break;
                        }
                        used[j]=0;
                    }
                }
            }
        }
        else puts("-1");
    }
    return 0;
}