| Time Limit: 6000MS | Memory Limit: 65536KB | 64bit IO Format: %lld & %llu |
Description
Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind, but not the dogs, so Jiajia use a special way to feed the dogs. At lunchtime, the dogs will stand on one line, numbered from 1 to n, the leftmost one is 1, the second one is 2, and so on. In each feeding, Jiajia choose an inteval[i,j], select the k-th pretty dog to feed. Of course Jiajia has his own way of deciding the pretty value of each dog. It should be noted that Jiajia do not want to feed any position too much, because it may cause some death of dogs. If so, Wind will be angry and the aftereffect will be serious. Hence any feeding inteval will not contain another completely, though the intervals may intersect with each other.
Your task is to help Jiajia calculate which dog ate the food after each feeding.
Your task is to help Jiajia calculate which dog ate the food after each feeding.
Input
The first line contains n and m, indicates the number of dogs and the number of feedings.
The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier.
Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding.
You can assume that n<100001 and m<50001.
The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier.
Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding.
You can assume that n<100001 and m<50001.
Output
Output file has m lines. The i-th line should contain the pretty value of the dog who got the food in the i-th feeding.
Sample Input
7 2
1 5 2 6 3 7 4
1 5 3
2 7 1
Sample Output
3
2
Source
POJ Monthly--2006.02.26,zgl & twb
区间求第k大值。
treap树维护第k大值。
用类似莫队的方法，将区间排序，每次添加新进区间的数，删去已经出区间的数，然后在树上找第k大数。
1 /*by SilverN*/ 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 #include<cstdio> 6 #include<cmath> 7 #include<cstdlib> 8 #include<ctime> 9 using namespace std; 10 const int mxn=100100; 11 int read(){ 12 int x=0,f=1;char ch=getchar(); 13 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 14 while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} 15 return x*f; 16 } 17 struct qry{ 18 int l,r; 19 int rk,id; 20 }a[mxn]; 21 int ans[mxn]; 22 int cmp(const qry q,const qry e){ 23 if(q.l!=e.l)return q.l<e.l; 24 return q.r<e.r; 25 } 26 struct node{ 27 int l,r; 28 int v,ct; 29 int size,rand; 30 }t[mxn]; 31 int root=0,cnt=0; 32 void update(int rt){ 33 t[rt].size=t[t[rt].l].size+t[t[rt].r].size+t[rt].ct; 34 return; 35 } 36 void ltt(int &rt){ 37 int now=t[rt].r; 38 t[rt].r=t[now].l; 39 t[now].l=rt; 40 t[now].size=t[rt].size; 41 update(rt); 42 rt=now; 43 return; 44 } 45 void rtt(int &rt){ 46 int now=t[rt].l; 47 t[rt].l=t[now].r; 48 t[now].r=rt; 49 t[now].size=t[rt].size; 50 update(rt); 51 rt=now; 52 return; 53 } 54 void insert(int &rt,int x){//新节点 插入值 55 if(!rt){ 56 t[++cnt].v=x; 57 t[cnt].size=1;t[cnt].rand=rand(); 58 t[cnt].ct=1; 59 rt=cnt; 60 return; 61 } 62 t[rt].size++; 63 if(t[rt].v==x)t[rt].ct++; 64 else if(x>t[rt].v){ 65 insert(t[rt].r,x); 66 if(t[t[rt].r].rand<t[rt].rand)ltt(rt); 67 } 68 else{ 69 insert(t[rt].l,x); 70 if(t[t[rt].l].rand<t[rt].rand)rtt(rt); 71 } 72 return; 73 } 74 void del(int &rt,int x){ 75 if(!rt)return; 76 if(t[rt].v==x){ 77 if(t[rt].ct>1){t[rt].size--;t[rt].ct--;return;} 78 if(t[rt].l*t[rt].r==0)rt=t[rt].l+t[rt].r;//子树补位 79 else{ 80 if(t[t[rt].l].rand<t[t[rt].r].rand){//把结点转到叶子上删除 81 rtt(rt); 82 del(rt,x); 83 } 84 else{ 85 ltt(rt); 86 del(rt,x); 87 } 88 } 89 return; 90 } 91 t[rt].size--; 92 if(x>t[rt].v)del(t[rt].r,x); 93 if(x<t[rt].v)del(t[rt].l,x); 94 return; 95 } 96 int query(int rt,int rank){ 97 if(!rt)return 0; 98 if(rank<=t[t[rt].l].size) return query(t[rt].l,rank); 99 if(rank>t[t[rt].l].size+t[rt].ct)return query(t[rt].r,rank-t[t[rt].l].size-t[rt].ct); 100 return t[rt].v; 101 } 102 // 103 int n,m; 104 int pt[mxn]; 105 106 int main(){ 107 t[0].size=0; 108 a[0].l=1;a[0].r=0; 109 srand(time(0)); 110 n=read();m=read(); 111 int i,j; 112 for(i=1;i<=n;i++)pt[i]=read(); 113 for(i=1;i<=m;i++){a[i].l=read();a[i].r=read();a[i].rk=read();a[i].id=i;} 114 sort(a+1,a+m+1,cmp); 115 for(i=1;i<=m;i++){ 116 if(a[i].l>a[i-1].r){ 117 root=0; 118 for(j=a[i].l;j<=a[i].r;j++)insert(root,pt[j]); 119 } 120 else{ 121 for(j=a[i-1].l;j<a[i].l;j++) del(root,pt[j]); 122 for(j=a[i-1].r+1;j<=a[i].r;j++) insert(root,pt[j]); 123 } 124 ans[a[i].id]=query(root,a[i].rk); 125 } 126 for(i=1;i<=m;i++)printf("%d ",ans[i]); 127 return 0; 128 }