Code Forces 652C Foe Pairs

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 ≤ 3·105)
— 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).

用一个数组表示每个数字可以向右延生的最大长度，也就是右边哪些点可以和这个数字形成一个区间。注意：

在给定完敌对点，更新数组之后，要从后往前再更新一次。相同左边端点的敌对点应该选择右端点较小的。

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
#define MAX 3*100000
int a[MAX+5];
int tag[MAX+5];
int dp[MAX+5];
int n,m;
int f[MAX+5];
int x,y;
int main()
{
    scanf("%d%d",&n,&m);
    memset(tag,0,sizeof(tag));
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        tag[a[i]]=i;
        f[i]=n;
    }
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&x,&y);
        int l=min(tag[x],tag[y]);
        int r=max(tag[x],tag[y]);
        f[l]=min(f[l],r-1);
    }

    for(int i=n-1;i>=1;i--)
    {
        f[i]=min(f[i],f[i+1]);
    }
    __int64 num=0;
    for(int i=1;i<=n;i++)
    {
        int right=f[i];
        num+=right-i+1;
    }
    printf("%I64d\n",num);
    return 0;

}