[APIO2018] Duathlon 铁人两项

题面

\(LOJ\)自己找。。

Sol

建立圆方树

考虑枚举起点\(s\)和终点\(t\)
那么答案就是\(s\)到\(t\)间的点双的点数和减去\(s,t\)
设方点权值为点双的点数，圆点的权值为\(-1\)
那么就是求\(s,t\)的路径上的点权和

现在考虑中间的点\(x\)
那么它的贡献就是经过它的路径的条数*它的权值
树\(DP\)得解

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;

IL int Input(){
	RG int x = 0, z = 1; RG char c = getchar();
	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	return x * z;
}

const int maxn(4e5 + 5);

int n, m, dfn[maxn], low[maxn], idx, sum;
int sta[maxn], top, tot, val[maxn], size[maxn];
ll ans;

struct Edge{
	int first[maxn], cnt, nxt[maxn << 1], to[maxn << 1];

	IL void Init(){
		cnt = 0, Fill(first, -1);
	}

	IL void Add(RG int u, RG int v){
		nxt[cnt] = first[u], to[cnt] = v, first[u] = cnt++;
	}
} e1, e2;

IL void Tarjan(RG int u){
	dfn[u] = low[u] = ++idx, sta[++top] = u, size[u] = 1;
	for(RG int e = e1.first[u]; e != -1; e = e1.nxt[e]){
		RG int v = e1.to[e];
		if(!dfn[v]){
			Tarjan(v), low[u] = min(low[u], low[v]);
			if(low[v] >= dfn[u]){
				RG int x = 0, num = 1; ++tot;
				do{
					x = sta[top--], ++num;
					e2.Add(tot, x), e2.Add(x, tot);
					size[tot] += size[x];
				} while(x != v);
				val[tot] = num, size[u] += size[tot];
				e2.Add(tot, u), e2.Add(u, tot);
			}
		}
		else low[u] = min(low[u], dfn[v]);
	}
}

IL void Dfs(RG int u, RG int ff){
	RG int tmp = u <= n;
	ans += 2LL * size[u] * (sum - size[u]) * val[u];
	for(RG int e = e2.first[u]; e != -1; e = e2.nxt[e]){
		RG int v = e2.to[e];
		if(v != ff){
			ans += 2LL * tmp * size[v] * val[u];
			tmp += size[v];
			Dfs(v, u);
		}
	}
}

int main(){
	e1.Init(), e2.Init();
	tot = n = Input(), m = Input();
	for(RG int i = 1; i <= n; ++i) val[i] = -1;
	for(RG int i = 1; i <= m; ++i){
		RG int u = Input(), v = Input();
		e1.Add(u, v), e1.Add(v, u);
	}
	for(RG int i = 1; i <= n; ++i)
		if(!dfn[i]) Tarjan(i), sum = size[i], Dfs(i, 0);
	printf("%lld\n", ans);
	return 0;
}