Educational Codeforces Round 10C Foe Pairs

最新推荐文章于 2022-08-17 23:26:38 发布

Dagon_

最新推荐文章于 2022-08-17 23:26:38 发布

阅读量331

点赞数

CC 4.0 BY-SA版权

分类专栏：脑洞题

本文链接：https://blog.youkuaiyun.com/q121999/article/details/78545447

脑洞题专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一种算法，用于解决给定排列和若干对数的情况下，如何计算不含这些对数的合法区间数量的问题。通过将对数转换为下标并排序，结合区间长度的枚举，实现了高效的解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Example
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).

求符合条件的区间的个数，要求区间内不能有列出的点对，给出的点对是数值，而区间是下标。
先枚举起点，再找最长区间长度。一开始想的是先枚举长度再枚举起点，太耗时间。
将点对转换成下标后按照右坐标从小到大排序，然后从左到右枚举起点，点对中左坐标在起点左边的直接排除，因为不可能被区间包含，所以排除后的第一个点对的右坐标就是区间的极限长度，由于已经包含了这个点对，更长的区间一定不符合条件。

#include <iostream>
#include <stdio.h>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <math.h>
#include <iterator>
#include <string.h>
using namespace std;
typedef long long ll;
int mo[4][2]={0,1,1,0,0,-1,-1,0};
const int MAXN=0x3f3f3f3f;
const int sz=600005;
int n,m;
int a[sz],pos[sz],vis[sz];
struct node{
    int len,l,r;
}edge[sz];
bool cmp(node x,node y){
    if(x.r==y.r){
        return x.l<y.l;
    }else{
        return x.r<y.r;
    }
}
ll ans;
void ss(){
    int j=1;
    for(int i=1;i<=n;i++){
        while(edge[j].l<i&&j<=m){
            j++;
        }
        if(j<=m)
            ans+=edge[j].r-i;
        else
            ans+=n+1-i;
        //cout<<i<<' '<<edge[j].r<<endl;
    }
}
int main()
{
    int x,y,len,pre,cot;
    //freopen("C:\\Users\\Administrator\\Desktop\\r.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            pos[a[i]]=i;
        }
        memset(vis,0,sizeof(vis));
        ans=0;
        for(int i=1;i<=m;i++){
            scanf("%d%d",&x,&y);
            x=pos[x];y=pos[y];
            if(x>y) swap(x,y);
            edge[i].l=x;
            edge[i].r=y;
        }
        sort(edge+1,edge+1+m,cmp);
        ss();
        printf("%I64d\n",ans);
    }
    return 0;
}