Educational Codeforces Round 10 C. Foe Pairs 水题-优快云博客

本文解析了 CodeForces C.FoePairs 题目，该题目要求找出不含特定敌对对组的区间数量。通过优化算法减少重复计算，实现了高效的求解。

C. Foe Pairs

题目连接：

http://www.codeforces.com/contest/652/problem/C

Description

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.

Sample Input

4 2
1 3 2 4
3 2
2 4

Sample Output

Hint

题意

给你一个n个数的排列

然后给你m组数

然后问你一共有多少个区间不包含这m组数。

题解：

对于每一个位置，我们直接暴力算出这个位置最多往左边延展多少。

但是显然这样子会有重复的，我们在扫的时候，维护一个当前最多往左边延展多少就好了

这样就不会有重复的了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e5+7;

int n,m,p[maxn],t[maxn];
int c[maxn];
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&p[i]);
        t[p[i]]=i;
    }
    for(int i=1;i<=m;i++)
    {
        int x,y;scanf("%d%d",&x,&y);
        int a = 0,b = 0;
        a=max(t[x],t[y]),b=min(t[x],t[y]);
        c[a]=max(c[a],b);
    }
    long long ans = 0;
    int l = 0;
    for(int i=1;i<=n;i++)
    {
        l=max(c[i],l);
        ans+=i-l;
    }
    cout<<ans<<endl;
}

转载于:https://www.cnblogs.com/qscqesze/p/5330279.html