2019杭电多校第二场hdu6601 Keen On Everything But Triangle

Keen On Everything But Triangle

题目传送门

解题思路

利用主席树求区间第k小，先求区间内最大的值，再求第二大，第三大……直到找到连续的三个数可以构成一个三角形。因为对于一组数，如果不能构成三角形，就小的就是斐波那契数列，因为数的范围在10^9内，所以不会超过50个数，也就是说，我们之间这样暴力地查询，查询次数不会超过50，肯定能找到结果。

代码如下

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

inline int read(){
    int res = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        res = (res << 3) + (res << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -res : res;
}

const int N = 100005;

int rt[N], cnt;
struct T{
    int l, r;
    int lch, rch;
    int sum;
}tree[N << 5];
ll a[N], b[N];
int build(int l, int r)
{
    int p = ++cnt;
    tree[p].l = l;
    tree[p].r = r;
    tree[p].lch = tree[p].rch = tree[p].sum = 0;
    if(l == r)
        return p;
    int mid = (l + r) / 2;
    tree[p].lch = build(l, mid);
    tree[p].rch = build(mid + 1, r);
    return p;
}

int insert(int k, int x)
{
    int p = ++cnt;
    tree[p].l = tree[k].l;
    tree[p].r = tree[k].r;
    tree[p].lch = tree[k].lch;
    tree[p].rch = tree[k].rch;
    tree[p].sum = tree[k].sum + 1;
    if(tree[k].l == tree[k].r)
        return p;
    int mid = (tree[k].l + tree[k].r) / 2;
    if(x <= mid)
        tree[p].lch = insert(tree[k].lch, x);
    else
        tree[p].rch = insert(tree[k].rch, x);
    return p;
}

int query(int k1, int k2, int x)
{
    if(tree[k1].l == tree[k1].r)
        return tree[k1].l;
    int l1 = tree[k1].lch;
    int l2 = tree[k2].lch;
    if(tree[l2].sum - tree[l1].sum >= x)
        return query(l1, l2, x);
    else {
        x -= tree[l2].sum - tree[l1].sum;
        return query(tree[k1].rch, tree[k2].rch, x);
    }
}

int main()
{
    int n, q;
    while(scanf("%d%d", &n, &q) != EOF){
        for(int i = 1; i <= n; i ++){
            scanf("%lld", &a[i]);
            b[i] = a[i];
        }
        sort(b + 1, b + n + 1);
        int k = unique(b + 1, b + n + 1) - b - 1;
        cnt = -1;
        rt[0] = build(1, k);
        for(int i = 1; i <= n; i ++){
            int x = lower_bound(b + 1, b + k + 1, a[i]) - b;
            rt[i] = insert(rt[i - 1], x);
        }
        for(int i = 1; i <= q; i ++){
            int l, r;
            l = read(), r = read();
            if(r - l + 1 < 3){
                printf("-1\n");
                continue;
            }
            int x = query(rt[l - 1], rt[r], r - l + 1);
            int y = query(rt[l - 1], rt[r], r - l);
            ll ans = -1;
            for(int i = r - l - 1; i >= 1; i --){
                int z = query(rt[l - 1], rt[r], i);
                if(b[y] + b[z] > b[x]){
                    ans = b[x] + b[y] + b[z];
                    break;
                }
                x = y;
                y = z;
            }
            printf("%lld\n", ans);
        }
    }
    return 0;
}