You’re given an undirected graph with n nodes and m edges. Nodes are numbered from 1 to n.

The graph is considered harmonious if and only if the following property holds:

For every triple of integers (l,m,r) such that 1≤l<m<r≤n, if there exists a path going from node l to node r, then there exists a path going from node l to node m.
In other words, in a harmonious graph, if from a node l we can reach a node r through edges (l<r), then we should able to reach nodes (l+1),(l+2),…,(r?1) too.

What is the minimum number of edges we need to add to make the graph harmonious?

Input
The first line contains two integers n and m (3≤n≤200 000 and 1≤m≤200 000).

The i-th of the next m lines contains two integers ui and vi (1≤ui,vi≤n, ui≠vi), that mean that there’s an edge between nodes u and v.

It is guaranteed that the given graph is simple (there is no self-loop, and there is at most one edge between every pair of nodes).

Output
Print the minimum number of edges we have to add to the graph to make it harmonious.

Examples
Input
14 8
1 2
2 7
3 4
6 3
5 7
3 8
6 8
11 12
Output
1
Input
200000 3
7 9
9 8
4 5
Output
0
Note
In the first example, the given graph is not harmonious (for instance, 1<6<7, node 1 can reach node 7 through the path 1→2→7, but node 1 can’t reach node 6). However adding the edge (2,4) is sufficient to make it harmonious.

In the second example, the given graph is already harmonious.
思路：我們利用并查集把相連通的點鏈接到一個集合內，祖先為最大的那個點。然后我們從1開始查找，設置_max為最大祖先，如果i<_max內出現了getf(i)!=_max的情況，就需要增加一條邊了，然后將兩個集合合并，最大祖先更新。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=2e5+100; int f[maxx]; int n,m;inline void init() {for(int i=0;i<=n+1;i++) f[i]=i; } inline int getf(int u) {return u==f[u]?u:f[u]=getf(f[u]); } inline void merge(int u,int v) {int t1=getf(u);int t2=getf(v);if(t1!=t2){if(t1>t2) f[t2]=t1;else f[t1]=t2;} } int main() {scanf("%d%d",&n,&m);init();int x,y;while(m--){scanf("%d%d",&x,&y);merge(x,y);}int ans=0;int _max=getf(1),tmp;for(int i=2;i<=n;i++){if(i<=_max){tmp=getf(i);if(tmp!=_max){merge(_max,tmp);_max=max(_max,tmp);ans++;}}else _max=getf(i);}cout<<ans<<endl;return 0; }

努力加油a啊，(^o)/~

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