hdu 6301 Distinct Values (2018多校 一 1004)(区间维护)

给出m个区间,每个区间内的数字不能有相等的,求字典序最小的结果序列.
排序区间后,维护区间端点,用set获得填的值.在遇到右端点(和当前队列顶部端点为同一区间)时补充set的值,这个过程是类似尺取的,只用n次.
注意细节即可.

#include <algorithm>
#include <cstring>
#include <iostream>
#include <string>
#include <queue>
#include <set>
#include <vector>
using namespace std;
#define debug(x) //std::cerr << #x << " = " << (x) << std::endl
typedef long long LL;
const int MAXN = 1e6 + 17;
const int MOD = 1e9 + 7;
struct dap
{
    int p,ty,id;
    dap(int a,int b,int c):p(a),ty(b),id(c){}
    bool operator< (const dap & b) const
    {
        if(p==b.p)
            return ty<b.ty;
        return p<b.p;
    }
};
int main(int argc, char const* argv[])
{
#ifdef noob
    freopen("Input.txt", "r", stdin);
    freopen("Output.txt", "w", stdout);
#endif
    int t;
    cin>>t;
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        vector<int > a(n+17),sur(n+17);
        vector<dap > all[n+17];
        set<int > s;
        for (int i = 0; i < m; ++i)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            sur[i] = 1;
            all[l].push_back(dap(l,-1,i));
            all[r].push_back(dap(r,1,i));
        }
        for (int i = 1; i <= n; ++i)
        {
            sort(all[i].begin(),all[i].end());
            s.insert(i);
        }
        bool in = false;
        queue<pair<int,int > > que;
        for (int i = 1; i <= n; ++i)
        {
            if(all[i].size())
            {
                a[i] = *s.begin();
                s.erase(a[i]);
                for (int j = 0; j < all[i].size(); ++j)
                {
                    int p = all[i][j].p,ty = all[i][j].ty,id = all[i][j].id;
                    if(ty==-1)
                        que.push({p,id});
                    else
                    {
                        if(!que.empty()&&que.front().second==id)
                        {
                            int st = que.front().first,ed = p;
                            bool nxt = false;
                            que.pop();
                            while(!que.empty())
                            {
                                if(sur[que.front().second])
                                {
                                    nxt = true;
                                    ed = que.front().first;
                                    break;
                                }
                                else
                                    que.pop();
                            }
                            debug(st);
                            if(nxt) ed--;
                            for (int k = st; k <= ed; ++k)
                                s.insert(a[k]);
                        }
                        else
                            sur[id] = 0;
                    }
                }
                if(que.size())
                    in = true;
                else
                    in = false;
            }
            else
            {
                if(!in) a[i] = 1;
                else
                {
                    a[i] = *s.begin();
                    s.erase(a[i]);
                }
            }
        }
        for (int i = 1; i <= n; ++i)
        {
            printf("%d%c", a[i],i==n?'\n':' ');
        }
    }
    return 0;
}