2019徐州网络赛I题

题目

Given a permutation pp of length nn, you are asked to answer mm queries, each query can be represented as a pair (l ,r )(l,r), you need to find the number of pair(i ,j)(i,j) such that l \le i < j \le rl≤i

Input

There is two integers n(1 \le n \le 10^5)n(1≤n≤105), m(1 \le m \le 10^5)m(1≤m≤105) in the first line, denoting the length of pp and the number of queries.

In the second line, there is a permutation of length nn, denoting the given permutation pp. It is guaranteed that pp is a permutation of length nn.

For the next mm lines, each line contains two integer l_ili​ and r_i(1 \le l_i \le r_i \le n)ri​(1≤li​≤ri​≤n), denoting each query.

Output

For each query, print a single line containing only one integer which denotes the number of pair(i,j)(i,j).

样例输入复制

3 2
1 2 3
1 3
2 3

样例输出复制

2
0

萌新终于学习到二维偏序这类题目的一些做法了。

#include 
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;
const int maxn=100010;
int id[maxn];
struct L
{
    int x,y;
}line[20*maxn];//要多开10倍
struct Q
{
    int l,r;
    int ind;
}q[maxn];
int a[maxn];
bool cmp1(L i,L j)
{

        return i.y>1;
    build(rt<<1,l,mid);
    build(rt<<1|1,mid+1,r);
}
void update(int rt,int p,int v)
{
    if(tree[rt].l==tree[rt].r)
    {
        tree[rt].va+=v;
        return;
    }
    int mid=(tree[rt].l+tree[rt].r)>>1;
    if(mid>=p) update(rt<<1,p,v);
    else update(rt<<1|1,p,v);
    pushup(rt);
}
int query(int rt,int L,int R)
{
    if(tree[rt].l>=L&&tree[rt].r<=R)
    {
        return tree[rt].va;
    }
    int mid=(tree[rt].l+tree[rt].r)>>1;
    int ans=0;
    if(mid>=L) ans+=query(rt<<1,L,R);
    if(midid[j])
            {
                swap(line[tot].x,line[tot].y);
            }
        }
    }
    for(int i=1;i<=m;i++)
    {
        scanf("%d %d",&q[i].l,&q[i].r);
        q[i].ind=i;
    }
    sort(line+1,line+tot+1,cmp1);
    sort(q+1,q+m+1,cmp2);
    int cnt=1;
    build(1,1,n);
    for(int i=1;i<=m;i++)
    {
        while(cnt<=tot&&line[cnt].y<=q[i].r)
        {
            update(1,line[cnt].x,1);
            cnt++;
        }
        res[q[i].ind]=query(1,q[i].l,q[i].r);
    }
    for(int i=1;i<=m;i++)
    {
        printf("%d\n",res[i]);
    }
    return 0;
}