codeforces 652C C. Foe Pairs(尺取法+线段树查询一个区间覆盖线段)

题目链接：

C. Foe Pairs

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

 

You are given a permutation p of length n. Also you are given m foe pairs (ai, bi) (1 ≤ ai, bi ≤ n, ai ≠ bi).

Your task is to count the number of different intervals (x, y) (1 ≤ x ≤ y ≤ n) that do not contain any foe pairs. So you shouldn't count intervals (x, y) that contain at least one foe pair in it (the positions and order of the values from the foe pair are not important).

Consider some example: p = [1, 3, 2, 4] and foe pairs are {(3, 2), (4, 2)}. The interval (1, 3) is incorrect because it contains a foe pair(3, 2). The interval (1, 4) is also incorrect because it contains two foe pairs (3, 2) and (4, 2). But the interval (1, 2) is correct because it doesn't contain any foe pair.

Input

 

The first line contains two integers n and m (1 ≤ n, m ≤ 3·10^5) — the length of the permutation p and the number of foe pairs.

The second line contains n distinct integers pi (1 ≤ pi ≤ n) — the elements of the permutation p.

Each of the next m lines contains two integers (ai, bi) (1 ≤ ai, bi ≤ n, ai ≠ bi) — the i-th foe pair. Note a foe pair can appear multiple times in the given list.

Output

 

Print the only integer c — the number of different intervals (x, y) that does not contain any foe pairs.

Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Examples

 

input

4 2
1 3 2 4
3 2
2 4

output

5

input

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

output

20

Note

In the first example the intervals from the answer are (1, 1), (1, 2), (2, 2), (3, 3) and (4, 4).

题意：

给一个数组,问有多少对(i,j)满足[a[i],a[j]]中不完整包含任何一个数对;

思路：

暴力是的复杂度太高,我先把这些数对都处理成原数组的位置,然后把它们搞进线段树里,就是把右端点当成左端点插入线段树时的位置,查询一个区间[i,j]时是否包含一个完整的线段可以看[i,j]中最大的左端点是多大,如果最大的左端点>=i时,我们就知道[i,j]中至少包含一个完整的线段;然后再枚举左端点,尺取法找右端点;然后把长度都加起来就是结果啦;

AC代码：

/*2014300227    652C - 11    GNU C++11    Accepted    311 ms    18796 KB*/
#include <bits/stdc++.h>
using namespace std;
const int N=3e5+;
typedef long long ll;
int n,a[N],fa[N],m,l,r;
struct Tree
{
    int l,r,ans;
};
Tree tree[*N];
void Pushup(int node)
{
    tree[node].ans=max(tree[*node].ans,tree[*node+].ans);
}
void build(int node,int L,int R)
{
    tree[node].l=L;
    tree[node].r=R;
    if(L==R)
    {
        tree[node].ans=;
        return ;
    }
    int mid=(L+R)>>;
    build(*node,L,mid);
    build(*node+,mid+,R);
    Pushup(node);
}
void update(int node,int num,int pos)
{
    if(tree[node].l==tree[node].r&&tree[node].r==pos)
    {
        tree[node].ans=max(tree[node].ans,num);
        return ;
    }
    int mid=(tree[node].l+tree[node].r)>>;
    if(pos<=mid)update(*node,num,pos);
    else update(*node+,num,pos);
    Pushup(node);
}
int query(int node,int L,int R)
{
    if(L<=tree[node].l&&R>=tree[node].r)
    {
        return tree[node].ans;
    }
    int mid=(tree[node].l+tree[node].r)>>;
    if(R<=mid)return query(*node,L,R);
    else if(L>mid)return query(*node+,L,R);
    else return max(query(*node,L,mid),query(*node+,mid+,R));
}
struct PO
{
    int l,r;
}po[N];
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=;i<=n;i++)
    {
        scanf("%d",&a[i]);
        fa[a[i]]=i;
    }
    build(,,n);
    for(int i=;i<m;i++)
    {
        scanf("%d%d",&l,&r);
        po[i].l=min(fa[l],fa[r]);
        po[i].r=max(fa[l],fa[r]);
        update(,po[i].l,po[i].r);//更新;
    }
    ll ans=;
    int l=,r=;
    for(int i=;i<=n;i++)
    {
        l=i;
        while(r<=n)
        {
            int q=query(,i,r);//查询[i,r]中的最大值,即是包含于这个区间的线段最大的左端点;
            if(q<i)r++;//r为以l为左端点满足要求的最长的区间的右端点+1;
            else break;
        }
        ans+=(ll)(r-l);
    }
    cout<<ans<<"\n";
    return ;
}