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 xi and has population pi, and all xi, as well as all pi, 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 xi and pi (1 ≤ xi,  pi ≤ 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

41 10007 109 112 100

Output

-1 4 2 1

Input

31 1002 13 10

Output

-1 1 1

Input

31 103 1002 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：

 1 #include <iostream>
 2 #include <queue>
 3 #include <stack>
 4 #include <cstdio>
 5 #include <vector>
 6 #include <map>
 7 #include <set>
 8 #include <bitset>
 9 #include <algorithm>
10 #include <cmath>
11 #include <cstring>
12 #include <cstdlib>
13 #include <string>
14 #include <sstream>
15 #define x first
16 #define y second
17 #define pb push_back
18 #define mp make_pair
19 #define lson l,m,rt*2
20 #define rson m+1,r,rt*2+1
21 #define mt(A,B) memset(A,B,sizeof(A))
22 #define mod 1000000007
23 using namespace std;
24 typedef long long LL;
25 const int N=200000+10;
26 const LL INF=0x3f3f3f3f3f3f3f3fLL;
27 LL val[N],id[N],x[N],ans[N],S[N],L[N],R[N];//S[N]表示单调栈
28 bool cmp(LL a,LL b)
29 {
30     return x[a]<x[b];
31 }
32 int main()
33 {
34 #ifdef Local
35     freopen("data.txt","r",stdin);
36 #endif
37     ios::sync_with_stdio(0);
38     LL i,j,k,n;
39     cin>>n;
40     for(i=1;i<=n;i++)
41     {
42         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
43         id[i]=i;
44     }
45     sort(id+1,id+1+n,cmp);//按在数轴上的位置从小到大排序
46     k=0;S[k++]=0;val[0]=2e9;//先把INF压入栈底
47     for(i=1;i<=n;i++)
48     {
49         while(val[S[k-1]]<=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
50         L[id[i]]=S[k-1];//找到了左边第一个比他大的数的下标
51         S[k++]=id[i];//把当前的下标压入栈
52     }
53     k=n+1;S[k--]=n+1;val[n+1]=2e9;
54     for(i=n;i>=1;i--)//同理
55     {
56         while(val[S[k+1]]<=val[id[i]])k++;
57         R[id[i]]=S[k+1];
58         S[k--]=id[i];
59     }
60     for(i=1;i<=n;i++)
61     {
62         if(L[i]==0&&R[i]==n+1)ans[i]=-1;
63         else if(L[i]==0)ans[i]=R[i];
64         else if(R[i]==n+1)ans[i]=L[i];
65         else
66         {
67             if(abs(x[L[i]]-x[i])<abs(x[R[i]]-x[i]))ans[i]=L[i];
68             else if(abs(x[L[i]]-x[i])>abs(x[R[i]]-x[i]))ans[i]=R[i];
69             else
70             {
71                 if(val[L[i]]<val[R[i]])ans[i]=R[i];
72                 else if(val[L[i]]>val[R[i]])ans[i]=L[i];
73             }
74         }
75     }
76     for(i=1;i<=n;i++)
77     {
78         if(i==n)cout<<ans[i]<<endl;
79         else cout<<ans[i]<<" ";
80     }
81 }



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：





  1 #include <iostream>
  2 #include <queue>
  3 #include <stack>
  4 #include <cstdio>
  5 #include <vector>
  6 #include <map>
  7 #include <set>
  8 #include <bitset>
  9 #include <algorithm>
 10 #include <cmath>
 11 #include <cstring>
 12 #include <cstdlib>
 13 #include <string>
 14 #include <sstream>
 15 #define lson l,m,rt*2
 16 #define rson m+1,r,rt*2+1
 17 #define mod 1000000007
 18 #define mt(A) memset(A,0,sizeof(A))
 19 using namespace std;
 20 typedef long long LL;
 21 const int N=200000+10;
 22 const LL INF=0x3f3f3f3f3f3f3f3fLL;
 23 LL dp[N][30];
 24 int ans[N];
 25 struct node
 26 {
 27     LL dis,val;
 28     int id;
 29 } a[N],b[N];
 30 bool cmp(struct node a,struct node b)
 31 {
 32     if(a.dis!=b.dis)return a.dis<b.dis;
 33     else return a.val>b.val;
 34 }
 35 void RMQ(int n)
 36 {
 37     for(int i=1; i<=n; i++)dp[i][0]=a[i].val;
 38     for(int j=1; (1<<j)<=n; j++)
 39     {
 40         for(int i=1; (i+(1<<j)-1)<=n; i++)
 41         {
 42             dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
 43         }
 44     }
 45 }
 46 LL query(int l,int r)
 47 {
 48     int k=0;
 49     while((1<<(k+1))<=(r-l+1))k++;
 50     return max(dp[l][k],dp[r-(1<<k)+1][k]);
 51 }
 52 struct node found(int l,int r,LL Val)//you
 53 {
 54     int mid;
 55     struct node p1;
 56     LL V;
 57     while(l<r)
 58     {
 59         mid=(l+r)/2;
 60         V=query(l,mid);
 61         if(V>Val)
 62         {
 63             r=mid;
 64         }
 65         else
 66         {
 67             l=mid+1;
 68         }
 69     }
 70     //cout<<l<<" "<<r<<endl;
 71     if(a[l].val>Val)p1=a[l];
 72     else p1.id=-1;
 73     return p1;
 74 }
 75 struct node found1(int l,int r,LL Val)//you
 76 {
 77     int mid;
 78     struct node p1;
 79     LL V;
 80     while(l<=r)
 81     {
 82         mid=(l+r)/2;
 83         V=query(mid,r);
 84         if(V>Val)
 85         {
 86             l=mid+1;
 87         }
 88         else
 89         {
 90             r=mid-1;
 91         }
 92     }
 93     //cout<<l<<" "<<r<<endl;
 94     if(a[r].val>Val)p1=a[r];
 95     else p1.id=-1;
 96     return p1;
 97 }
 98 int main()
 99 {
100 #ifdef Local
101     freopen("data","r",stdin);
102 #endif
103     int i,j,k,n;
104     cin>>n;
105     for(i=1; i<=n; i++)
106     {
107         scanf("%lld%lld",&a[i].dis,&a[i].val);
108         a[i].id=i;
109         b[i]=a[i];
110     }
111     sort(a+1,a+n+1,cmp);
112     RMQ(n);
113     for(i=1; i<=n; i++)
114     {
115         int l,r,mid;
116         struct node p1,p2;
117         p1.id=-1;p2.id=-1;
118         LL V;
119         l=1;
120         r=i;
121         p1=found1(l,r,a[i].val);
122         l=i;r=n;
123         p2=found(l,r,a[i].val);
124         //cout<<i<<" "<<p1.id<<" "<<p2.id<<endl;
125         if(p2.id==-1&&p1.id==-1)ans[a[i].id]=-1;
126         else if(p1.id==-1&&p2.id!=-1)ans[a[i].id]=p2.id;
127         else if(p1.id!=-1&&p2.id==-1)ans[a[i].id]=p1.id;
128         else
129         {
130             if(abs(a[i].dis-p1.dis)>abs(a[i].dis-p2.dis))
131             {
132                 ans[a[i].id]=p2.id;
133             }
134             else if(abs(a[i].dis-p1.dis)<abs(a[i].dis-p2.dis))
135             {
136                 ans[a[i].id]=p1.id;
137             }
138             else
139             {
140                 if(p1.val>p2.val)ans[a[i].id]=p1.id;
141                 else ans[a[i].id]=p2.id;
142             }
143         }
144     }
145     for(i=1;i<=n;i++)
146     {
147         if(i==n)printf("%d\n",ans[i]);
148         else printf("%d ",ans[i]);
149     }
150 }






				


				
时间： 2024-10-05 21:31:57 

		


		


	

	
	

		


		
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