[CF310]D. Case of Fugitive

题意：
给出n个线段，在n个线段之间搭桥，给出m个桥的长度，假如满足条件
To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i?+?1)-th islads, if there are such coordinates of x and y, that li?≤?x?≤?ri, li?+?1?≤?y?≤?ri?+?1 and y?-?x?=?a.
，问最后可以连接所有的线段吗？是的话输出线段所对应的桥的编号。

分析：
数据量比较大，虽然题意很简单，但是要想一些办法降低复杂度。这个数据量二分可以nlogn可以处理，那这里就用了set来处理，二分就用了lower_bound，另外题一点，pair

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <utility> 
#include<stack>
#include <algorithm>
#define read freopen("q.in","r",stdin)
#define LL long long
#define maxn 200005
using namespace std;

set <pair<pair<LL,LL>, int> > s;
multiset<pair<LL,LL> > k;
int res[maxn];
int main()
{ //‘std::set<std::pair<std::pair<long long int, long long int>, int> >::iterator‘ has no member named ‘first‘
    LL n,m,x,y,a,b;
    int i,j;
    std::ios::sync_with_stdio(false);
    cin>>n>>m;

    for(i=0;i<n;i++)
    {
        cin>>x>>y;
        if(i>0)s.insert(make_pair(make_pair(y-a,x-b),i));
        a=x;b=y;
    }
    for(i=1;i<=m;i++)
    {
        cin>>x;
        k.insert(make_pair(x,i));
    }
    for(set<pair<pair<LL,LL>, int> >:: iterator it =s.begin();it!=s.end();it++)
    {
        multiset<pair<LL,LL> > :: iterator  j=k.lower_bound(make_pair(it->first.second,-1));
        if(j==k.end() || j->first>it->first.first) 
        {
            cout<<"No"<<endl;
            return 0;
        }
        res[it->second]=j->second;
        k.erase(j);
    }
    cout<<"Yes"<<endl;
    for(i=1;i<n;i++)cout<<res[i]<<" ";
    cout<<endl;
}

ps:觉得对STL不熟。。。