Atcoder2134 Zigzag MST

weixin_30756499

于 2019-07-14 13:17:00 发布

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原文链接：http://www.cnblogs.com/LSlzf/p/11183991.html

博客围绕有N个顶点的图展开，要处理Q次加边查询。介绍了输入输出格式及样例，解析中提出等价替换思路，将边进行替换使图产生环，用类似最大前缀方式求f值，更新两圈后，图中剩q+n条边，最后用Kruskal求最大生成树。

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 Ai,Bi 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 Ai+1 will be connected by an edge with a weight of Ci+1.
The two vertices numbered Ai+1 and Bi+1 will be connected by an edge with a weight of *C**i*+2.
The two vertices numbered Bi+1 and Ai+2 will be connected by an edge with a weight of *C**i*+3.
The two vertices numbered Ai+2 and Bi+2 will be connected by an edge with a weight of *C**i*+4.
The two vertices numbered Bi+2 and Ai+3 will be connected by an edge with a weight of *C**i*+5.
The two vertices numbered Ai+3 and Bi+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;
}

转载于:https://www.cnblogs.com/LSlzf/p/11183991.html