洛谷P4630 [APIO2018] Duathlon 铁人两项 【圆方树】

题目链接

洛谷P4630

题解

看了一下部分分，觉得树的部分很可做，就相当于求一个点对路径长之和的东西，考虑一下能不能转化到一般图来？
一般图要转为树，就使用圆方树呗

思考一下发现，两点之间经过的点双，点双内所有点一定都可以作为中介点
那么我们将方点赋值为点双大小，为了去重，剩余点赋值(-1)
答案就是任意两点间权值和之和

我们只需枚举每个点被经过多少次，这就很容易计算了

复杂度(O(n))

#include<algorithm>
#include<iostream>
#include<cstdio>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define LL long long int
using namespace std;
const int maxn = 400005,maxm =1000005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int hh[maxn],Ne = 1,h[maxn],ne = 1;
struct EDGE{int from,to,nxt;}e[maxm],ed[maxm];
inline void build(int u,int v){
	e[++Ne] = (EDGE){u,v,hh[u]}; hh[u] = Ne;
	e[++Ne] = (EDGE){v,u,hh[v]}; hh[v] = Ne;
}
inline void add(int u,int v){
	ed[++ne] = (EDGE){u,v,h[u]}; h[u] = ne;
	ed[++ne] = (EDGE){v,u,h[v]}; h[v] = ne;
}
int n,m,val[maxn],dfn[maxn],low[maxn],inst[maxn],st[maxm],top,cnt,N;
void combine(int rt){
	val[++N] = 1;
	add(N,rt);
	while (st[top] != rt && low[st[top]] >= dfn[rt]){
		inst[st[top]] = false;
		add(N,st[top--]),val[N]++;
	}
}
void dfs(int u,int s){
	dfn[u] = low[u] = ++cnt; st[++top] = u; inst[u] = true;
	for (int k = hh[u],to; k; k = e[k].nxt){
		if (k == s) continue;
		if (!dfn[to = e[k].to]){
			dfs(to,k ^ 1);
			if (low[to] > dfn[u]){
				top--; val[++N] = 2;
				add(u,N); add(N,to);
			}
			else if (low[to] == dfn[u]) combine(u);
			low[u] = min(low[u],low[to]);
		}
		else if (inst[to]) low[u] = min(low[u],dfn[to]);
	}
}
int fa[maxn];
LL siz[maxn],ans,sum;
void DFS(int u){
	siz[u] = u <= n ? 1 : 0;
	Redge(u) if ((to = ed[k].to) != fa[u]){
		fa[to] = u,DFS(to),siz[u] += siz[to];
	}
	LL tot = u <= n ? sum - 1 : 0;
	Redge(u) if ((to = ed[k].to) != fa[u]){
		tot += siz[to] * (sum - siz[to]);
	}
	tot += (sum - siz[u]) * siz[u];
	ans += tot * val[u];
}
int main(){
	N = n = read(); m = read();
	REP(i,m) build(read(),read());
	REP(i,n) val[i] = -1;
	REP(i,n) if (!dfn[i]){
		sum = cnt; top = 0; dfs(i,-1);
		sum = cnt - sum; DFS(i);
	}
	printf("%lld
",ans);
	return 0;
}

总结

1、圆方树的写法
2、点双的写法
3、一般图考虑树的做法的时候，可以考虑建圆方树