Educational Codeforces Round 10 C. Foe Pairs —— 后缀和

最新推荐文章于 2023-08-16 16:37:13 发布

原创最新推荐文章于 2023-08-16 16:37:13 发布 · 536 阅读

0 ·

CC 4.0 BY-SA版权

前缀和 & 后缀和专栏收录该内容

5 篇文章

订阅专栏

本文详细解析 CodeForces 平台上的 652C 题目，介绍如何通过记录敌人对的位置来计算不包含敌人对的有效区间数量，并给出具体的实现代码。

题目链接：http://codeforces.com/problemset/problem/652/C

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).

题解：

1.pos[val]记录val的下标，lim[l]记录以l为左端点，其敌人作为右端点的最小下标。

2.为使得每个间隔不出现死敌，所以需要对lim[l]进行修改，做法是从右往左更新lim[l]的最小值。

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 3e5+10;

int n, m;
int pos[maxn], lim[maxn];

void init()
{
    scanf("%d%d",&n,&m);
    for(int i = 1; i<=n; i++)
    {
        int v;
        scanf("%d",&v);
        pos[v] = i;
        lim[i] = n + 1;
    }
}

void solve()
{
    for(int i = 1; i<=m; i++)
    {
        int x, y;
        scanf("%d%d",&x,&y);
        int l = pos[x];
        int r = pos[y];
        if(l>r) swap(l,r);
        lim[l] = min(lim[l], r);
    }

    for(int i = n-1; i>=1; i--)
        lim[i] = min(lim[i], lim[i+1]);

    LL ans = 0;
    for(int i = 1; i<=n; i++)
        ans += 1LL*lim[i] - 1LL*i;
    cout<<ans<<endl;
}

int main()
{
    init();
    solve();
}