[ABC232G] Modulo Shortest Path

Problem Statement

We have a directed graph with $N$ vertices, called Vertex $1$, Vertex $2$, $\ldots$, Vertex $N$.

For each pair of integers such that $1 \leq i, j \leq N$ and $i \neq j$, there is a directed edge from Vertex $i$ to Vertex $j$ of weight $(A_i + B_j) \bmod M$. (Here, $x \bmod y$ denotes the remainder when $x$ is divided by $y$.)

There is no edge other than the above.

Print the shortest distance from Vertex $1$ to Vertex $N$, that is, the minimum possible total weight of the edges in a path from Vertex $1$ to Vertex $N$.

Constraints

• $2 \leq N \leq 2 \times 10^5$
• $2 \leq M \leq 10^9$
• $0 \leq A_i, B_j < M$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$N$ $M$
$A_1$ $A_2$ $\ldots$ $A_N$
$B_1$ $B_2$ $\ldots$ $B_N$

Output

Print the minimum possible total weight of the edges in a path from Vertex $1$ to Vertex $N$.

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Sample Input 1

4 12
10 11 6 0
8 7 4 1

Sample Output 1

3

Below, $i \rightarrow j$ denotes the directed edge from Vertex $i$ to Vertex $j$.

Let us consider the path $1$ $\rightarrow$ $3$ $\rightarrow$ $2$ $\rightarrow$ $4$.

• Edge $1\rightarrow 3$ weighs $(A_1 + B_3) \bmod M = (10 + 4) \bmod 12 = 2$,
• Edge $3 \rightarrow 2$ weighs $(A_3 + B_2) \bmod M = (6 + 7) \bmod 12 = 1$,
• Edge $2\rightarrow 4$ weighs $(A_2 + B_4) \bmod M = (11 + 1) \bmod 12 = 0$.

Thus, the total weight of the edges in this path is $2 + 1 + 0 = 3$.

This is the minimum possible sum for a path from Vertex $1$ to Vertex $N$.

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Sample Input 2

10 1000
785 934 671 520 794 168 586 667 411 332
363 763 40 425 524 311 139 875 548 198

Sample Output 2

462

一道优化建图题.

观察到这里的取模最多只会减一次，所以与其说是取模，不如说当 \(a_i+b_j\ge m\) 时，代价减去 \(m\)。

先看如果没有取模，怎么做。一个经典的拆点，把一个点拆成内点和外点，外点向内点连代价为 \(b_i\) 的边，内点向外点连 \(a_i\) 的边。然后随便用0边把外点连起来。

但这样很明显不好扩展。注意到一件事，如果 \(a_i+b_j\ge m\)，将点按照 \(b\) 排序后，\(k>j\) 的点的代价都要减去 \(m\)。所以将所有点按照 \(b\) 排序，然后仍然拆成内点和外点，外点从点 \(i\) 连向 \(i+1\),代价 \(b_{i+1}-b_i\)。内外点之间连 0 边。这个构图非常巧妙，如果从点 \(i\) 的内点向某一个点连了一条边权为 \(a_i+b_j-m\) 的边，这个内点到达任何一个 \(k>j\)，都相当于有一条边权为 \(a_i+b_k-m\) 的边。对于一个内点，他先朝 \(1\) 的外点连一条边权为 \(a_i+b_1\) 的边，在二分出第一个 \(a_i+b_j\ge m\) 的点，连一条边权为 \(a_i+b_j-m\) 的边，就可以达到题目中的效果。

但这样好像有些点同时连了 \(a_i+b_j\) 和 \(a_i+b_j-m\) 的边？但其实不影响答案。因为题目求最短路。

所有的边权为正，可以跑dij.

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=4e5+5;
int n,b[N],e_num,m,u,v,hd[N],vis[N];
LL dis[N];
struct node{
	int a,b,id;
	bool operator<(const node&n)const{
		return b<n.b;
	}
}a[N];
struct edge{
	int v,nxt,w;
}e[N<<3];
struct dian{
	int v;
	LL w;
	bool operator<(const dian&d)const{
		return w>d.w;
	}
};
priority_queue<dian>q;
void add_edge(int u,int v,int w)
{
//	printf("%d %d %d\n",u,v,w);
	e[++e_num]=(edge){v,hd[u],w};
	hd[u]=e_num;
}
void dijkstra(int s)
{
	q.push((dian){s,0});
	memset(dis,0x7f,sizeof(dis));
	dis[s]=0;
	while(!q.empty())
	{
		int k=q.top().v;
		q.pop();
		if(vis[k])
			continue;
		vis[k]=1;
		for(int i=hd[k];i;i=e[i].nxt)
		{
			if(dis[e[i].v]>dis[k]+e[i].w)
			{
				dis[e[i].v]=dis[k]+e[i].w;
				q.push((dian){e[i].v,dis[e[i].v]});
			}
		}
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i].a);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i].b),a[i].id=i;
	sort(a+1,a+n+1);
	for(int i=1;i<=n;i++)
		b[i]=a[i].b;
	for(int i=1;i<=n;i++)
	{
//		b[i]=a[i].b;
		if(a[i].id==1)
			u=i;
		else if(a[i].id==n)
			v=i;
		add_edge(i+n,i,0);
		add_edge(i,n+1,b[1]+a[i].a);
		int k=lower_bound(b+1,b+n+1,m-a[i].a)-b;
//		printf("%d %d\n",k,a[i].a);
		if(k<=n)
			add_edge(i,k+n,a[i].a+b[k]-m);
	}
//	printf("%d %d\n",u,v);
	for(int i=1;i<n;i++)
		add_edge(i+n,i+n+1,b[i+1]-b[i]);
	dijkstra(u);
//	for(int i=1;i<=n+n;i++)
//		printf("%lld ",dis[i]);
	printf("%lld",dis[v]);
	return 0;
}