杭电多校11题

http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1011&cid=849
就会俩题。。。10题水题不写了，今天还做了两道主席树板题也不写了。。。
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

给一个数组，每次查询l到r区间能组成的最大周长三角形
在区间内从大到小遍历一次即可
比赛的时候感觉主席树时空复杂度太大就没写。。。然后就白给了
重新开数组或者用优先队列每次查询都是O（nlogn）
主席树也是，但是查询时不一定遍历完所以时间复杂度会更小

#include
#include
#include
#define maxl 100005
using namespace std;
int n,q;
int nn,tot;
int rt[maxl];
long long a[maxl],b[maxl];
struct node
{
	int ls,rs,sum;
}tree[maxl*30];
void insert(int num,int &x,int l,int r)
{
	tree[++tot]=tree[x];
	x=tot;
	++tree[x].sum;
	if(l==r) return;
	int mid=(l+r)>>1;
	if(num<=mid)
		insert(num,tree[x].ls,l,mid);
	else
		insert(num,tree[x].rs,mid+1,r);
}

void prework()
{
	tree[0].ls=tree[0].rs=tree[0].sum=0;
	rt[0]=0;
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
		b[i]=a[i];
	} 
	sort(b+1,b+1+n);
	nn=unique(b+1,b+n+1)-b-1;
	for(int i=1;i<=n;i++)
	a[i]=lower_bound(b+1,b+nn+1,a[i])-b;
	tot=0;
	for(int i=1;i<=n;i++)
	{
		rt[i]=rt[i-1];
		insert(a[i],rt[i],1,nn);
	}
}

int query(int i,int j,int k,int l,int r)
{
	if(l==r)
		return l;
	int tp=tree[tree[j].ls].sum-tree[tree[i].ls].sum;
	int mid=(l+r)>>1;
	if(k<=tp) 
		return query(tree[i].ls,tree[j].ls,k,l,mid);
	else
		return query(tree[i].rs,tree[j].rs,k-tp,mid+1,r);
}
int main()
{
	while(~scanf("%d%d",&n,&q))
	{
		tot=0;
		prework();
		while(q--)
		{
			int L,R;
			scanf("%d%d",&L,&R);
			if(R-L<2)
			{
				printf("-1\n");
				continue;
			}
			long long a1,a2,a3;
			a1=b[query(rt[L-1],rt[R],R-L+1,1,nn)];
			a2=b[query(rt[L-1],rt[R],R-L,1,nn)];
			long long ans=-1;
			for(int i=R-L-1;i>=1;i--)
			{			
				a3=b[query(rt[L-1],rt[R],i,1,nn)];
				if(a1<(a2+a3))
				{
					ans=a1+a2+a3;
					break;
				}
				a1=a2;
				a2=a3;
			}			
		    printf("%lld\n",ans);
		}	
	}
}