Atcoder2134 Zigzag MST

问题描述

We have a graph with N vertices, numbered 0 through N−1. Edges are yet to be added.

We will process Q queries to add edges. In the i-th (1≦i≦Q) query, three integers A**i,B**i and C**i will be given, and we will add infinitely many edges to the graph as follows:

The two vertices numbered A**i and B**i will be connected by an edge with a weight of C**i.
The two vertices numbered B**i and A**i+1 will be connected by an edge with a weight of C**i+1.
The two vertices numbered A**i+1 and B**i+1 will be connected by an edge with a weight of C**i+2.
The two vertices numbered B**i+1 and A**i+2 will be connected by an edge with a weight of C**i+3.
The two vertices numbered A**i+2 and B**i+2 will be connected by an edge with a weight of C**i+4.
The two vertices numbered B**i+2 and A**i+3 will be connected by an edge with a weight of C**i+5.
The two vertices numbered A**i+3 and B**i+3 will be connected by an edge with a weight of C**i+6.
...

Here, consider the indices of the vertices modulo N. For example, the vertice numbered N is the one numbered 0, and the vertice numbered 2N−1 is the one numbered N−1.

输入格式

The input is given from Standard Input in the following format:

N Q

A1 B1 C1

A2 B2 C2

:

AQ BQ CQ

输出格式

Print the total weight of the edges contained in a minimum spanning tree of the graph.

样例输入

7 1

5 2 1

样例输出

解析

考虑等价替换。对于一次加边操作，由于 (x,y) 的权值比 (y,x+1) 的权值小，考虑kruskal 的过程，在考虑边 (y,x+1,z+1) 的时候，x,y 一定已在一个连通块里。于是，我们可以将 (y,x+1,z+1) 等价替换为 (x,x+1,z+1)。同理，(x+1,y+1,z+2) 可替换为(y,y+1,z+2)。经过这样的替换后，我们发现，一次加边操作会使图上产生两个环，以a为起点的环的路径为c+1,c+3,c+5,......，以b为起点的环的路径为c+2,c+4,c+6,......，同时还有(a,b,c)的边。设\(f[i]\)表示从i-1到i之间的边中最小的，我们可以用类似于最大前缀的方式，求出每一个f:

\[f[i+1]=max(f[i+1],f[i])
\]

注意\(f[n+1]=f[1]\)，因为在更新一圈后可能还有可以被后来的点更新的，所以我们要更新两圈。

最后，图中只剩下q+n条边，用Kruskal求最大生成树即可。

代码

#include <iostream>

#include <cstdio>

#include <algorithm>

#include <cstring>

#define int long long

#define N 400002

using namespace std;

struct edge{

	int u,v,w;

}e[N];

int n,q,i,f[N],num,fa[N];

int read()

{

	char c=getchar();

	int w=0;

	while(c<'0'||c>'9') c=getchar();

	while(c<='9'&&c>='0'){

		w=w*10+c-'0';

		c=getchar();

	}

	return w;

}

int my_comp(const edge &x,const edge &y)

{

	return x.w<y.w;

}

int find(int x)

{

	if(x!=fa[x]) fa[x]=find(fa[x]);

	return fa[x];

}

int Kruscal()

{

	for(int i=1;i<=n;i++) fa[i]=i;

	sort(e+1,e+num+1,my_comp);

	int cnt=n,ans=0;

	for(int i=1;i<=num;i++){

		int f1=find(e[i].u),f2=find(e[i].v);

		if(f1!=f2){

			fa[f1]=f2;

			cnt--;

			ans+=e[i].w;

		}

	}

	return ans;

}

signed main()

{

	memset(f,0x3f,sizeof(f));

	n=read();q=read();

	for(i=1;i<=q;i++){

		int a,b,c;

		a=read()+1;b=read()+1;c=read();

		e[++num]=(edge){a,b,c};

		f[a]=min(f[a],c+1);

		f[b]=min(f[b],c+2);

	}

	bool flag=0;

	for(i=1;i<=n;i++){

		f[i%n+1]=min(f[i]+2,f[i%n+1]);

		if(i==n&&!flag) i=0,flag=1;

	}

	for(i=1;i<=n;i++) e[++num]=(edge){i,i%n+1,f[i]};

	cout<<Kruscal()<<endl;

	return 0;

}

巴特西