\(\text{Solution}\)

一个性质：两个 \(K\) 元组有边相连当且仅当每个点在对应的图中到 \(1\) 有奇偶性相同的路径
那么我们就可以预处理每个图中的点到 \(1\) 的奇偶最短路
再考虑路径长度，显然是 \(\min(\max_{i=1}^k{odd_i}, \max_{i=1}^k{even_i})\)
把它拆开 \(\max_{i=1}^k{odd_i} + \max_{i=1}^k{even_i} - \max_{i=1}^k(\max(odd_i,even_i))\)

那么一个点的信息就成了三个
分别算出这三个 \(\max\) 即可
先枚举 \(\max\) 即最大的是多少，那么 \(K\) 元组其他位置就是其他图中小于 \(\max\) 的点的个数乘起来
可以从大枚举 \(\max\)，把点信息为 \(\max\) 先算了，再删去这个点即可，那么就用线段树维护区间积即可，线段树下标表示第 \(i\) 个图剩余点的个数
好处是线段树中剩余的点信息一定小于等于当前枚举的 \(\max\)，避免既要处理图不同又要处理信息大小关系的困境

\(\text{Code}\)

#include <cstdio> 
#include <queue>
#include <vector>
#include <iostream>
#include <cstring>
#define re register
#define LL long long
using namespace std;

const int K = 5e4 + 5, N = 2e5 + 5, INF = 0x3f3f3f3f;
LL MOD = 1e9 + 7;
int mx, k, n, m, h[N], tot, total, dis[N][2], col[N], vis[N][2];
vector<int> g[N][3];
struct node{int x, z;};
queue<node> Q;

struct edge{int to, nxt;}e[N << 1];
inline void add(int x, int y)
{
	e[++tot] = edge{y, h[x]}, h[x] = tot;
	e[++tot] = edge{x, h[y]}, h[y] = tot;
}

inline void read(int &x)
{
	x = 0; char ch = getchar();
	while (!isdigit(ch)) ch = getchar();
	while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
}

struct Segment{
	#define ls (p << 1)
	#define rs (ls | 1)
	int tr[K << 2]; LL sum[K << 2];
	void build(int p, int l, int r)
	{
		if (l == r) return void(sum[p] = 0);
		int mid = (l + r) >> 1;
		build(ls, l, mid), build(rs, mid + 1, r);
	}
	void update(int p, int l, int r, int x, int v)
	{
		if (x < l || x > r) return;
		if (l == r) return void(sum[p] += v);
		int mid = (l + r) >> 1;
		if (x <= mid) update(ls, l, mid, x, v);
		else update(rs, mid + 1, r, x, v);
		sum[p] = sum[ls] * sum[rs] % MOD;
	}
	LL query(int p, int l, int r, int tl, int tr)
	{
		if (tl > r || tr < l) return 1;
		if (tl <= l && r <= tr) return sum[p];
		int mid = (l + r) >> 1;
		LL res = 1;
		if (tl <= mid) res = query(ls, l, mid, tl, tr);
		if (tr > mid) res = res * query(rs, mid + 1, r, tl, tr) % MOD;
		return res;
	}
}T;

inline void spfa()
{
	for(re int j = 1; j <= n; j++) dis[j][0] = dis[j][1] = INF, vis[j][0] = vis[j][1] = 0;
	dis[1][0] = 0, vis[1][0] = 1, Q.push(node{1, 0});
	while (!Q.empty())
	{
		node now = Q.front(); Q.pop();
		for(re int j = h[now.x]; j; j = e[j].nxt)
		if (dis[e[j].to][now.z ^ 1] > dis[now.x][now.z] + 1)
		{
			dis[e[j].to][now.z ^ 1] = dis[now.x][now.z] + 1;
			if (!vis[e[j].to][now.z ^ 1])
				vis[e[j].to][now.z ^ 1] = 1, Q.push(node{e[j].to, now.z ^ 1});
		}
		vis[now.x][now.z] = 0;
	}
}

inline LL solve(int z)
{
	T.build(1, 1, k);
	for(re int i = mx; i >= 0; i--)
		for(re int j = 0; j < g[i][z].size(); j++) T.update(1, 1, k, col[g[i][z][j]], 1);
	LL res = 0;
	for(re int i = mx; i; i--)
		for(re int j = 0; j < g[i][z].size(); j++)
		{
			int now = g[i][z][j];
			res = (res + T.query(1, 1, k, 1, col[now] - 1) * T.query(1, 1, k, col[now] + 1, k) % MOD * i % MOD) % MOD;
			T.update(1, 1, k, col[now], -1);
		}
	return res;
}

int main()
{
	read(k);
	for(re int i = 1; i <= k; i++)
	{
		tot = 0;
		read(n), read(m);
		for(re int j = 1, x, y; j <= m; j++) read(x), read(y), add(x, y);
		spfa();
		for(re int j = 1; j <= n; j++)
		{
			col[total + j] = i;
			for(re int l = 0; l <= 1; l++)
			if (dis[j][l] ^ INF) g[dis[j][l]][l].push_back(total + j);
			if (max(dis[j][0], dis[j][1]) ^ INF) g[max(dis[j][0], dis[j][1])][2].push_back(total + j);
			if (dis[j][0] ^ INF) mx = max(mx, dis[j][0]);
			if (dis[j][1] ^ INF) mx = max(mx, dis[j][1]);
		}
		total += n;
		for(re int i = 1; i <= n; i++) h[i] = 0;
	}
	printf("%lld\n", (solve(0) + solve(1) - solve(2) + MOD) % MOD);
}

当然用 \(STL\) 的 \(vector\) 和 \(queue\) 虽然直观但终究是慢了

\(\text{Code}\)

#include <cstdio> 
#include <iostream>
#define re register
#define LL long long
using namespace std;

const int K = 5e4 + 5, N = 2e5 + 5, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int mx, k, n, m, h[N], tot, total, dis[N][2], col[N], vis[N][2], g[N][3], Tot;
struct node{int x, z;};
node Q[N << 1];
struct edge{int to, nxt;}e[N << 1], E[N << 1];
inline void add(int x, int y)
{
	e[++tot] = edge{y, h[x]}, h[x] = tot;
	e[++tot] = edge{x, h[y]}, h[y] = tot;
}
inline void Add(int x, int z, int y){E[++Tot] = edge{y, g[x][z]}, g[x][z] = Tot;}

inline void read(int &x)
{
	x = 0; char ch = getchar();
	while (!isdigit(ch)) ch = getchar();
	while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
}

struct Segment{
	#define ls (p << 1)
	#define rs (ls | 1)
	int tr[K << 2], sum[K << 2];
	void build(int p, int l, int r)
	{
		if (l == r) return void(sum[p] = 0);
		int mid = (l + r) >> 1;
		build(ls, l, mid), build(rs, mid + 1, r);
	}
	void update(int p, int l, int r, int x, int v)
	{
		if (x < l || x > r) return;
		if (l == r) return void(sum[p] += v);
		int mid = (l + r) >> 1;
		if (x <= mid) update(ls, l, mid, x, v);
		else update(rs, mid + 1, r, x, v);
		sum[p] = 1LL * sum[ls] * sum[rs] % MOD;
	}
	int query(int p, int l, int r, int tl, int tr)
	{
		if (tl > r || tr < l) return 1;
		if (tl <= l && r <= tr) return sum[p];
		int mid = (l + r) >> 1;
		int res = 1;
		if (tl <= mid) res = query(ls, l, mid, tl, tr);
		if (tr > mid) res = 1LL * res * query(rs, mid + 1, r, tl, tr) % MOD;
		return res;
	}
}T;

inline void spfa()
{
	int head = 0, tail = 1;
	for(re int j = 1; j <= n; j++) dis[j][0] = dis[j][1] = INF, vis[j][0] = vis[j][1] = 0;
	dis[1][0] = 0, vis[1][0] = 1, Q[1] = node{1, 0};
	while (head < tail)
	{
		node now = Q[++head];
		for(re int j = h[now.x]; j; j = e[j].nxt)
		if (dis[e[j].to][now.z ^ 1] > dis[now.x][now.z] + 1)
		{
			dis[e[j].to][now.z ^ 1] = dis[now.x][now.z] + 1;
			if (!vis[e[j].to][now.z ^ 1])
				vis[e[j].to][now.z ^ 1] = 1, Q[++tail] = node{e[j].to, now.z ^ 1};
		}
		vis[now.x][now.z] = 0;
	}
}

inline int solve(int z)
{
	T.build(1, 1, k);
	for(re int i = mx; i >= 0; i--)
		for(re int j = g[i][z]; j; j = E[j].nxt) T.update(1, 1, k, col[E[j].to], 1);
	int res = 0;
	for(re int i = mx; i; i--)
		for(re int j = g[i][z]; j; j = E[j].nxt)
			res = (res + 1LL * T.query(1, 1, k, 1, col[E[j].to] - 1) * T.query(1, 1, k, col[E[j].to] + 1, k) % MOD * i % MOD) % MOD,
			T.update(1, 1, k, col[E[j].to], -1);
	return res;
}

int main()
{
	read(k);
	for(re int i = 1; i <= k; i++)
	{
		tot = 0;
		read(n), read(m);
		for(re int j = 1, x, y; j <= m; j++) read(x), read(y), add(x, y);
		spfa();
		for(re int j = 1; j <= n; j++)
		{
			col[total + j] = i;
			for(re int l = 0; l <= 1; l++)
			if (dis[j][l] ^ INF) Add(dis[j][l], l, total + j);
			if (max(dis[j][0], dis[j][1]) ^ INF) Add(max(dis[j][0], dis[j][1]), 2, total + j);
			if (dis[j][0] ^ INF) mx = max(mx, dis[j][0]);
			if (dis[j][1] ^ INF) mx = max(mx, dis[j][1]);
		}
		total += n;
		for(re int i = 1; i <= n; i++) h[i] = 0;
	}
	printf("%d\n", (1LL * solve(0) + solve(1) - solve(2) + MOD) % MOD);
}

【USACO 2021 January Contest, Platinum】Problem 1. Sum of Distances