LibreOJ #516. 「LibreOJ β Round #2」DP 一般看规律

二次联通门 : LibreOJ #516. 「LibreOJ β Round #2」DP 一般看规律

 

 

 

 

/*
    LibreOJ #516. 「LibreOJ β Round #2」DP 一般看规律
    
    set启发式合并
    
    题目中给定的代码意思求相同的数中间隔最小的值。
    
    那么维护n个set就好
    合并时把小的向大的上暴力合并
    
    用了map所以不用离散化 
*/
#include <iostream>
#include <cstdio>
#include <set>
#include <map>

#define INF 2147483647

void read (int &now)
{
    register char word = getchar ();
    int temp = 0;
    for (; !isdigit (word); word = getchar ())
        if (word == '-')
            temp = 1;
    for (now = 0; isdigit (word); now = now * 10 + word - '0', word = getchar ());
    if (temp)
        now = -now;
}

std :: map <int, std :: set <int> > Need;
int N, M;
int Answer = INF;

inline int min (int a, int b)
{
    return a < b ? a : b;
}

inline void Insert (int pos, int x)
{
    std :: set <int> :: iterator now = Need[pos].lower_bound (x);
    if (now != Need[pos].end ())
        Answer = min (Answer, *now - x);
    if (now != Need[pos].begin ())
        Answer = min (Answer, x - *-- now);
    Need[pos].insert (x);
}

int main (int argc, char *argv[])
{
    read (N), read (M);
    int x, y;
    for (int i = 1; i <= N; i ++)
    {
        read (x);
        Insert (x, i);
    }
    
    for (; M; -- M)
    {
        read (x), read (y);
        
        if (x != y)
        {
            if (Need[x].size () > Need[y].size ())
                std :: swap (Need[x], Need[y]);
            for (std :: set <int> :: iterator i = Need[x].begin (); i != Need[x].end (); ++ i)
                Insert (y, *i);
            Need[x].clear ();
        }
        printf ("%d\n", Answer);
    }
    
    return 0;
}