题意描述

You are given an array consisting of n integers a1, a2, …, an. Initially ax=1, all other elements are equal to 0.

You have to perform m operations. During the i-th operation, you choose two indices c and d such that li≤c,d≤ri, and swap ac and ad.

Calculate the number of indices k such that it is possible to choose the operations so that ak=1 in the end.

每次可以交换两个数，询问1的最多个数

思路

我们可以维护最小左边界和最大的右边界，最初区间内只有一个点x，然后遍历m个操作，如果区间可以合并，则更新边界即可。

AC代码

#include<bits/stdc++.h>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long ll;
const int N=205;
const int M=1e6+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;
struct node{
    int l,r;
}Node[N];
void solve(){
    int n,m,x;cin>>n>>x>>m;
    rep(i,0,m) cin>>Node[i].l>>Node[i].r;
    int max_r=x,min_l=x;
    rep(i,0,m){
        int l=Node[i].l,r=Node[i].r;
        if(l<=max_r && r>=min_l){
            max_r=max(max_r,r);
            min_l=min(min_l,l);
        }
    }
    cout<<max_r-min_l+1<<endl;
}
int main(){
    IOS;
    int t;cin>>t;
    while(t--){
        solve();
    }
    return 0;
}