hdu 5324 Boring Class （树套树）

Boring Class

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 255    Accepted Submission(s): 53

Problem Description

Mr. Zstu and Mr. Hdu are taking a boring class , Mr. Zstu comes up with a problem to kill time, Mr. Hdu thinks it’s too easy, he solved it very quickly, what about you guys?
Here is the problem:
Give you two sequences  L1,L2,...,Ln and  R1,R2,...,Rn.
Your task is to find a longest subsequence  v1,v2,...vm satisfies
v1≥1, vm≤n, vi<vi+1 .(for i from 1 to m - 1)
Lvi≥Lvi+1, Rvi≤Rvi+1(for i from 1 to m - 1)
If there are many longest subsequence satisfy the condition, output the sequence which has the smallest lexicographic order.

 

Input

There are several test cases, each test case begins with an integer n.
1≤n≤50000
Both of the following two lines contain n integers describe the two sequences.
1≤Li,Ri≤109

 

Output

For each test case ,output the an integer m indicates the length of the longest subsequence as described.
Output m integers in the next line.

 

Sample Input

5 5 4 3 2 1 6 7 8 9 10 2 1 2 3 4

 

Sample Output

5 1 2 3 4 5 1 1

 

Source

2015 Multi-University Training Contest 3

 

这道题是经典的树套树，外面用树状数组，里面套Treap树。一维树状数组节点先要对数据离散化，第二维Treap节点维护一个mlen和midx统计所有该节点左孩子、右孩子和自己DP最优值。空间是动态分配的，由于n次查找，每次对树状数组上的logn个一维结点查询，每次最多往每个一维结点中插入一个二维Treap结点，空间的总开销为n*logn，时间是n*logn*logn。

从数组的右边往左边做DP，可以保证字典序最小。时间比较慢1500ms。

    #include 
    #include 
    #include 
    #include 
    #include 

    using namespace std;

    const int maxn = 50010;

    struct Node{
        Node *ch[2];
        int r;
        int v; //第二维
        int len; //长度
        int idx; //下标
        int mlen; //结点和结点左孩子右孩子里最大长度
        int midx; //结点和结点左孩子右孩子里最大长度对应的下标
        Node() {}
        Node(int v, int len, int idx):v(v),len(len),idx(idx),mlen(len),midx(idx) {ch[0] = ch[1] = NULL; r = rand();}

        int cmp(int x) const {
            if (x == v) return -1;
            return x < v ? 0 : 1;
        }

        void maintain(){

            mlen = len;
            midx = idx;

            if (ch[0] != NULL && (ch[0]->mlen > mlen || (ch[0]->mlen == mlen && ch[0]->midx < midx))){
                mlen = ch[0]->mlen;
                midx = ch[0]->midx;
            }

            if (ch[1] != NULL && (ch[1]->mlen > mlen || (ch[1]->mlen == mlen && ch[1]->midx < midx))){
                mlen = ch[1]->mlen;
                midx = ch[1]->midx;
            }
        }
    };

    bool findMax(Node* a, Node* b){
        if (a->mlen < b->mlen || (a->mlen == b->mlen && a->midx > b->midx)){
            *a = *b;
            return true;
        }
        return false;
    }

    namespace Treap{

        int cntnode;

        Node node[maxn*10];

        void init(){
            cntnode = 0;
        }

        Node* newNode(int v, int len, int idx){

            node[++cntnode] = Node(v, len, idx);
            return &node[cntnode];
        }

        void rotate(Node* &o, int d){

            Node* k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o;
            o->maintain(); k->maintain(); o = k;
        }

        void insert(Node* &o, int v, int len, int idx){

            if (o == NULL) o = newNode(v, len, idx);
            else {
                int d = o->cmp(v);
                if (d != -1){
                    insert(o->ch[d], v, len, idx);
                    if (o->r < o->ch[d]->r) rotate(o, d^1);
                }
                else
                {
                    if (len >= o->len){
                        o->len = len;
                        o->idx = idx;
                    }
                }
            }
            o->maintain();
        }

        Node search(Node *o, int v){
            if (o == NULL){
                return Node(-1, 0, -1);
            }
            else{
                Node re, tmp;
                if (o->v == v) {
                    re = Node(o->v, o->len, o->idx);
                    if (o->ch[1]){
                        findMax(&re, o->ch[1]);
                    }
                }
                else if (o->v > v){
                    re = Node(o->v, o->len, o->idx);
                    if (o->ch[1]){
                        findMax(&re, o->ch[1]);
                    }
                    if (o->ch[0]){
                        tmp = search(o->ch[0], v);
                        findMax(&re, &tmp);
                    }
                }
                else{
                    re = search(o->ch[1], v);
                }
                return re;
            }
        }

    }

    namespace BIT{

        Node* fwt[maxn];
        int N;

        void init(int n){
            N = n;
            memset(fwt, 0, sizeof fwt);
        }

        void add(int v1, int v2, int len, int idx){
            while(v1 < N){
                Treap::insert(fwt[v1], v2, len, idx);
                v1 += (-v1)&v1;
            }
        }

        Node query(int v1, int v2){
            Node re, tmp;
            re = Node(-1, 0, -1);
            while(v1 > 0){
                tmp = Treap::search(fwt[v1], v2);
                findMax(&re, &tmp);
                v1 -= (-v1)&v1;
            }
            return re;
        }

    }

    struct Pe{
        int L,R;
        int i;
        bool operator < (const Pe& rhs)const{
            return L < rhs.L;
        }
    };

    bool cmp(Pe a, Pe b){
        return a.i < b.i;
    }

    int solo[maxn];
    Pe pe[maxn];int pre[maxn];

    void print(Node& a){
        int id = a.midx;
        printf("%d\n", a.mlen);
        while(1){
            printf("%d", id+1);
            if (pre[id] == -1){
                break;
            }
            printf(" ");
            id = pre[id];
        }
        printf("\n");
    }

    int main(){

        int n;
        while(scanf("%d", &n) != EOF){

            for(int i=0;i=0;i--){
                tmp = BIT::query(pe[i].L, pe[i].R);
                pre[i] = tmp.midx;
                tmp.mlen = tmp.mlen + 1;
                tmp.midx = i;
                BIT::add(pe[i].L, pe[i].R, tmp.mlen, tmp.midx);
                findMax(&ans, &tmp);
            }

            print(ans);
        }

        return 0;
    }