[loj 2587] 「APIO2018」铁人两项

「APIO2018」铁人两项

LOJ 2587

题目大意

给出一个 \(n\) 个点，\(m\) 条边的无向图，问存在多少个互异三元组 \((s, c, f)\) 满足存在两条分别从 \(s\) 到 \(c\) ，\(c\) 到 \(f\) ，且点不重复的路径

数据范围

\(n \le 100000, m \le 200000\)

时空限制

1000ms, 1024MB

分析

圆方树模板题

考虑对于一对 \((s, f)\) ，所有可能的 \(c\) 就是它们路径上点双里的点，而且不算 \(s\) ，\(f\) ，建出圆方树，考虑路径上一个点双的贡献是它的大小，而割点会重复统计，所以让圆方树上圆点贡献为 \(-1\) ，方点贡献为它的 \(size\) 那么一个点的计算次数就是通过它的圆点对数，树形DP即可

总结

最开始想的是枚举 \(c\) 算贡献，所以发现不可做时要更换思路啊...

Code

#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>

using namespace std;

inline char nc()
{
    static char buf[100000], * l = buf, * r = buf;
    if (l == r) r = (l = buf) + fread(buf, 1, 100000, stdin);
    if (l == r) return EOF;
    return *l++;
}
template<class T> void read(T & x)
{
    x = 0; int f = 1, ch = nc();
    while (!isdigit(ch))
    {
        if (ch == '-') f = -1;
        ch = nc();
    }
    while (isdigit(ch))
    {
        x = x * 10 - '0' + ch;
        ch = nc();
    }
    x *= f;
}

typedef long long ll;

const int maxn = 100000 + 5;
const int maxm = 200000 + 5;
const int maxe = maxm * 2;
const int maxnode = maxn * 2;

int n, m;

struct edge
{
    int to, nex;
    edge(int to = 0, int nex = 0) : to(to), nex(nex) {}
} g[maxe];
int head[maxn];
int ecnt;
vector<int> adj[maxnode];

inline void addedge(int u, int v)
{
    g[ecnt] = edge(v, head[u]), head[u] = ecnt++;
    g[ecnt] = edge(u, head[v]), head[v] = ecnt++;
}
inline void adde(int u, int v)
{
    adj[u].push_back(v);
    adj[v].push_back(u);
}

int dfc, dfn[maxn], low[maxn];
int top, sta[maxn];
int bcnt;
int bccsiz[maxn];

int val[maxnode];
int siz[maxnode];

void tarjan(int u, int fa)
{
    low[u] = dfn[u] = ++dfc;
    sta[++top] = u;
    siz[u] = 1;
    for(int i = head[u]; ~ i; i = g[i].nex)
    {
        int v = g[i].to;
        if(v != fa)
        {
            if(!dfn[v])
            {
                tarjan(v, u);
                low[u] = min(low[u], low[v]);
                if(low[v] >= dfn[u])
                {
                    bcnt++;
                    adde(n + bcnt, u);
                    bccsiz[bcnt] = 1;
                    while(true)
                    {
                        int x = sta[top--];
                        adde(n + bcnt, x);
                        bccsiz[bcnt]++;
                        siz[bcnt + n] += siz[x];
                        if(x == v) break;
                    }
                    siz[u] += siz[bcnt + n];
                    val[bcnt + n] = bccsiz[bcnt];
                }
            }
            else low[u] = min(low[u], dfn[v]);
        }
    }
}

int all;
ll an;

void dfs(int u, int fa)
{
    ll re = (ll)(all - siz[u]) * siz[u] * 2;
    int t = int(u <= n);
    for(unsigned int i = 0; i < adj[u].size(); ++i)
    {
        int v = adj[u][i];
        if(v != fa)
        {
            dfs(v, u);
            re += (ll)t * siz[v] * 2;
            t += siz[v];
        }
    }
    an += re * val[u];
}

void solve()
{
    for(int i = 1; i <= n; ++i)
    {
        val[i] = -1;
    }
    for(int i = 1; i <= n; ++i) if(!dfn[i])
    {
        tarjan(i, -1);
        all = siz[i], dfs(i, -1);
    }
    cout << an << endl;
}

int main()
{
//    freopen("testdata.in", "r", stdin);
    read(n), read(m);
    memset(head, -1, sizeof(head));
    for(int i = 1; i <= m; ++i)
    {
        int u, v; read(u), read(v);
        addedge(u, v);
    }
    solve();
    return 0;
}