Code Forces Gym 100971D Laying Cables（单调栈）

D - Laying Cables

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

One-dimensional country has n cities, the i-th of which is located at the point x_i and has population p_i, and all x_i, as well as all p_i, are distinct. When one-dimensional country got the Internet, it was decided to place the main server in the largest city, and to connect any other city j to the city k that has bigger population than j and is the closest to it (if there are many such cities, the largest one should be chosen). City k is called a parent of city j in this case.

Unfortunately, the Ministry of Communications got stuck in determining from where and to where the Internet cables should be laid, and the population of the country is suffering. So you should solve the problem. For every city, find its parent city.

Input

The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of cities.

Each of the next n lines contains two space-separated integers x_i and p_i (1 ≤ x_i,  p_i ≤ 10⁹) — the coordinate and the population of thei-th city.

Output

Output n space-separated integers. The i-th number should be the parent of the i-th city, or  - 1, if the i-th city doesn't have a parent. The cities are numbered from 1 by their order in the input.

Sample Input

Input

4
1 1000
7 10
9 1
12 100

Output

-1 4 2 1

Input

3
1 100
2 1
3 10

Output

-1 1 1

Input

3
1 10
3 100
2 1

Output

2 -1 2

题意：给定一维数轴上的N个点和对应的权值，同时定义一个点的父亲为：离它最近且权值比它大的点下标，如果距离相同则选择权值较大的，如果没有则为-1，求出所有点的父亲？

思路1：单调栈预处理出左边和右边第一个比它大的元素的下标即可。
       单调栈的作用：求出某个数的左边或右边第一个比它大或比它小的元素，复杂度O(n)。
       单调递增栈：元素从栈底到栈顶严格单调递增。
       单调递减栈：元素从栈底到栈顶严格单调递减。

代码1：

 #include <iostream>
 #include <queue>
 #include <stack>
 #include <cstdio>
 #include <vector>
 #include <map>
 #include <set>
 #include <bitset>
 #include <algorithm>
 #include <cmath>
 #include <cstring>
 #include <cstdlib>
 #include <string>
 #include <sstream>
 #define x first
 #define y second
 #define pb push_back
 #define mp make_pair
 #define lson l,m,rt*2
 #define rson m+1,r,rt*2+1
 #define mt(A,B) memset(A,B,sizeof(A))
 #define mod 1000000007
 using namespace std;
 typedef long long LL;
 const int N=+;
 const LL INF=0x3f3f3f3f3f3f3f3fLL;
 LL val[N],id[N],x[N],ans[N],S[N],L[N],R[N];//S[N]表示单调栈
 bool cmp(LL a,LL b)
 {
     return x[a]<x[b];
 }
 int main()
 {
 #ifdef Local
     freopen("data.txt","r",stdin);
 #endif
     ios::sync_with_stdio();
     LL i,j,k,n;
     cin>>n;
     for(i=;i<=n;i++)
     {
         cin>>x[i]>>val[i];//x[i]表示在数轴上的位置，val[i]表示该点的价值
         id[i]=i;
     }
     sort(id+,id++n,cmp);//按在数轴上的位置从小到大排序
     k=;S[k++]=;val[]=2e9;//先把INF压入栈底
     for(i=;i<=n;i++)
     {
         while(val[S[k-]]<=val[id[i]])k--;//如果当前的val比栈顶的大，就把栈顶的元素弹出来
         L[id[i]]=S[k-];//找到了左边第一个比他大的数的下标
         S[k++]=id[i];//把当前的下标压入栈
     }
     k=n+;S[k--]=n+;val[n+]=2e9;
     for(i=n;i>=;i--)//同理
     {
         while(val[S[k+]]<=val[id[i]])k++;
         R[id[i]]=S[k+];
         S[k--]=id[i];
     }
     for(i=;i<=n;i++)
     {
         if(L[i]==&&R[i]==n+)ans[i]=-;
         else if(L[i]==)ans[i]=R[i];
         else if(R[i]==n+)ans[i]=L[i];
         else
         {
             if(abs(x[L[i]]-x[i])<abs(x[R[i]]-x[i]))ans[i]=L[i];
             else if(abs(x[L[i]]-x[i])>abs(x[R[i]]-x[i]))ans[i]=R[i];
             else
             {
                 if(val[L[i]]<val[R[i]])ans[i]=R[i];
                 else if(val[L[i]]>val[R[i]])ans[i]=L[i];
             }
         }
     }
     for(i=;i<=n;i++)
     {
         if(i==n)cout<<ans[i]<<endl;
         else cout<<ans[i]<<" ";
     }
 }

思路2：先按数轴上的位置排序，RMQ预处理区间的最大值，再往左往右二分。训练的时候并不会单调栈，所以只能用RMQ二分瞎几把乱搞。

代码2：

 #include <iostream>
 #include <queue>
 #include <stack>
 #include <cstdio>
 #include <vector>
 #include <map>
 #include <set>
 #include <bitset>
 #include <algorithm>
 #include <cmath>
 #include <cstring>
 #include <cstdlib>
 #include <string>
 #include <sstream>
 #define lson l,m,rt*2
 #define rson m+1,r,rt*2+1
 #define mod 1000000007
 #define mt(A) memset(A,0,sizeof(A))
 using namespace std;
 typedef long long LL;
 const int N=+;
 const LL INF=0x3f3f3f3f3f3f3f3fLL;
 LL dp[N][];
 int ans[N];
 struct node
 {
     LL dis,val;
     int id;
 } a[N],b[N];
 bool cmp(struct node a,struct node b)
 {
     if(a.dis!=b.dis)return a.dis<b.dis;
     else return a.val>b.val;
 }
 void RMQ(int n)
 {
     for(int i=; i<=n; i++)dp[i][]=a[i].val;
     for(int j=; (<<j)<=n; j++)
     {
         for(int i=; (i+(<<j)-)<=n; i++)
         {
             dp[i][j]=max(dp[i][j-],dp[i+(<<(j-))][j-]);
         }
     }
 }
 LL query(int l,int r)
 {
     int k=;
     while((<<(k+))<=(r-l+))k++;
     return max(dp[l][k],dp[r-(<<k)+][k]);
 }
 struct node found(int l,int r,LL Val)//you
 {
     int mid;
     struct node p1;
     LL V;
     while(l<r)
     {
         mid=(l+r)/;
         V=query(l,mid);
         if(V>Val)
         {
             r=mid;
         }
         else
         {
             l=mid+;
         }
     }
     //cout<<l<<" "<<r<<endl;
     if(a[l].val>Val)p1=a[l];
     else p1.id=-;
     return p1;
 }
 struct node found1(int l,int r,LL Val)//you
 {
     int mid;
     struct node p1;
     LL V;
     while(l<=r)
     {
         mid=(l+r)/;
         V=query(mid,r);
         if(V>Val)
         {
             l=mid+;
         }
         else
         {
             r=mid-;
         }
     }
     //cout<<l<<" "<<r<<endl;
     if(a[r].val>Val)p1=a[r];
     else p1.id=-;
     return p1;
 }
 int main()
 {
 #ifdef Local
     freopen("data","r",stdin);
 #endif
     int i,j,k,n;
     cin>>n;
     for(i=; i<=n; i++)
     {
         scanf("%lld%lld",&a[i].dis,&a[i].val);
         a[i].id=i;
         b[i]=a[i];
     }
     sort(a+,a+n+,cmp);
     RMQ(n);
     for(i=; i<=n; i++)
     {
         int l,r,mid;
         struct node p1,p2;
         p1.id=-;p2.id=-;
         LL V;
         l=;
         r=i;
         p1=found1(l,r,a[i].val);
         l=i;r=n;
         p2=found(l,r,a[i].val);
         //cout<<i<<" "<<p1.id<<" "<<p2.id<<endl;
         if(p2.id==-&&p1.id==-)ans[a[i].id]=-;
         else if(p1.id==-&&p2.id!=-)ans[a[i].id]=p2.id;
         else if(p1.id!=-&&p2.id==-)ans[a[i].id]=p1.id;
         else
         {
             if(abs(a[i].dis-p1.dis)>abs(a[i].dis-p2.dis))
             {
                 ans[a[i].id]=p2.id;
             }
             else if(abs(a[i].dis-p1.dis)<abs(a[i].dis-p2.dis))
             {
                 ans[a[i].id]=p1.id;
             }
             else
             {
                 if(p1.val>p2.val)ans[a[i].id]=p1.id;
                 else ans[a[i].id]=p2.id;
             }
         }
     }
     for(i=;i<=n;i++)
     {
         if(i==n)printf("%d\n",ans[i]);
         else printf("%d ",ans[i]);
     }
 }