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
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：
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 ",ans[i]); 148 else printf("%d ",ans[i]); 149 } 150 }