HDU 6601(2019杭电多校二 1011) Keen On Everything But Triangle（主席树 + 组成三角形相关）

Keen On Everything But Triangle

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0
 

Problem Description

N sticks are arranged in a row, and their lengths are a1,a2,...,aN.

There are Q querys. For i-th of them, you can only use sticks between li-th to ri-th. Please output the maximum circumference of all the triangles that you can make with these sticks, or print −1 denoting no triangles you can make.

 

Input

There are multiple test cases.

Each case starts with a line containing two positive integers N,Q(N,Q≤105).

The second line contains N integers, the i-th integer ai(1≤ai≤109) of them showing the length of the i-th stick.

Then follow Q lines. i-th of them contains two integers li,ri(1≤li≤ri≤N), meaning that you can only use sticks between li-th to ri-th.

It is guaranteed that the sum of Ns and the sum of Qs in all test cases are both no larger than 4×105.

 

Output

For each test case, output Q lines, each containing an integer denoting the maximum circumference.

 

Sample Input

5 3 2 5 6 5 2 1 3 2 4 2 5

 

Sample Output

13 16 16

 

题意：给定 n 根木棍的长度，然后有 Q 个询问，每一次询问给定 l , r 表示询问从第 l 根木棍到第 r 根木棍中能够组成的 三角形的最大周长。。

思路：关于用木棍组成三角形的问题，你需要知道当木棍数量 >= 47 根时一定能组成三角形（极端情况是 47 根木棍的长度成斐波拉契数，这些数无法构成三角形，且容纳的边数最多。第 47 项超过1e9），所以对于最大周长，一定在前 47 根木棍中构成。。所以每次询问利用主席树暴力 获得 前 47 根木棍长度，然后算周长即可。时间复杂度  Q * 50 * logn

Code:

#include
#define debug(x) cout << "[" << #x <<": " << (x) <<"]"<< endl
#define pii pair
#define clr(a,b) memset((a),b,sizeof(a))
#define rep(i,a,b) for(int i = a;i < b;i ++)
#define pb push_back
#define MP make_pair
#define LL long long
#define ull unsigned LL
#define ls i << 1
#define rs (i << 1) + 1
#define INT(t) int t; scanf("%d",&t)

using namespace std;

const int maxn = 1e5 + 10;
const int M = maxn * 40;
int T[maxn];///T[i]表示第i颗线段树的顶结点
int lson[M],rson[M],c[M];///C[i]就代表第i个点的权值,即上述权值线段树的sum
int X[maxn],K[maxn];     /// lson[i], rson[i] 表示第 i 个点的左儿子,右儿子是谁
int tot = 0,en;          /// tot 用于动态开点

int getId(int x){
    return lower_bound(X + 1,X + 1 + en,x) - X;
}

int build(int l,int r){
    int rt = tot ++;
    c[rt] = 0;
    if(l != r){
        int mid = (l + r) >> 1;
        lson[rt] = build(l,mid);
        rson[rt] = build(mid + 1,r);
    }
    return rt;
}

int update(int rt,int pos,int val){
    int newrt = tot ++,tmp = newrt;
    c[newrt] = c[rt] + val;
    int l = 1,r = en;
    while(l < r){
        int mid = (l + r) >> 1;
        if(pos <= mid){
            lson[newrt] = tot ++;
            rson[newrt] = rson[rt];
            newrt = lson[newrt];
            rt = lson[rt];
            r = mid;
        }
        else {
            rson[newrt] = tot ++;
            lson[newrt] = lson[rt];
            newrt = rson[newrt];
            rt = rson[rt];
            l = mid + 1;
        }
        c[newrt] = c[rt] + val;
    }
    return tmp;
}

int query(int lrt,int rrt,int k){
    int l = 1,r = en;
    while(l < r){
        int mid = (l + r) >> 1;
        if(c[rson[rrt]] - c[rson[lrt]] >= k){   ///简单画一画就理解了
            l = mid + 1;
            lrt = rson[lrt];
            rrt = rson[rrt];
        }
        else {
            r = mid;
            k -= c[rson[rrt]] - c[rson[lrt]];
            rrt = lson[rrt];
            lrt = lson[lrt];
        }
    }
    return l;
}

LL LEN[60];
int p = 0;

int main() {
    int n,q;
    while(~scanf("%d%d",&n,&q)){
        en = tot = 0;
        for(int i = 1;i <= n;++ i){
            scanf("%d",&K[i]);
            X[++ en] = K[i];
        }
        sort(X + 1,X + 1 + en);
        en = unique(X + 1,X + 1 + en) - X - 1;
        T[0] = build(1,en);
        for(int i = 1;i <= n;++ i){
            T[i] = update(T[i - 1],getId(K[i]),1);
        }
        while(q --){
            int l,r; scanf("%d%d",&l,&r);
            p = 0;
            for(int j = 1;j <= min(50,r - l + 1);++ j){
                LEN[++ p] = X[query(T[l - 1],T[r],j)];
            }
//            for(int j = 1;j <= p;++ j) printf("%lld ",LEN[j]); printf("\n");
            LL flag = 0;
            for(int j = 1;j + 2 <= p;++ j){
                if(LEN[j] < LEN[j + 1] + LEN[j + 2]){
                    flag = LEN[j] + LEN[j + 1] + LEN[j + 2];
                    break;
                }
            }
            if(!flag) printf("-1\n");
            else printf("%lld\n",flag);
        }
    }
    return 0;
}