【LOJ2587】【APIO2018】铁人两项（圆方树，树形DP）

Description

https://loj.ac/problem/2587

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Solution

发现其实题目要求的就是对于每一条路径$(u,v)$所有可能经过的点数的和。

我们建出圆方树，令方点的权值为所在点双（一条边也看作点双）的大小，圆点权值为$-1$，那么路径$(u,v)$可能经过的点的个数就是圆方树上两点间路径的权值和。

我们要计算对于每一条路径的和，树形DP一下就行了。

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Code

/**************************************
 * Au: Hany01
 * Prob: [LOJ2587][APIO2018] 铁人两项
 * Date: Jul 31st, 2018
 * Email: [email protected]
**************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = j; i < i##_end_; ++ i)
#define For(i, j ,k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define SZ(a) ((int)(a.size()))
#define ALL(a) a.begin(), a.end()
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define x first
#define y second
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define y1 wozenmezhemecaia 
#ifdef hany01
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...)
#endif

template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
    register char c_; register int _, __;
    for (_ = 0, __ = 1, c_ = getchar(); !isdigit(c_); c_ = getchar()) if (c_ == '-')  __ = -1;
    for ( ; isdigit(c_); c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 1e5 + 5, maxm = maxn << 2;

int n, e = 1, beg1[maxn], beg2[maxn << 1], nex[maxm << 1], v[maxm << 1], sz[maxn << 2], dfn[maxn], low[maxn], clk, w[maxn << 1], tot, all, stk[maxn], top, cnt;
LL  Ans;

inline void add1(int uu, int vv) { v[++ e] = vv, nex[e] = beg1[uu], beg1[uu] = e; }
inline void add2(int uu, int vv) { v[++ e] = vv, nex[e] = beg2[uu], beg2[uu] = e; }

void DFS(int u, int fe) {
    dfn[u] = low[u] = ++ clk, stk[++ top] = u, w[u] = -1;
    for (register int i = beg1[u]; i; i = nex[i]) if ((i >> 1) != fe) {
        if (!dfn[v[i]]) {
            DFS(v[i], i >> 1), chkmin(low[u], low[v[i]]);
            if (low[v[i]] >= dfn[u]) {
                add2(u, ++ tot), cnt = 1;
                do add2(tot, stk[top]), ++ cnt; while (stk[top --] != v[i]);
                w[tot] = cnt;
            }
        } else chkmin(low[u], dfn[v[i]]);
    }
}

void getsz(int u) {
    sz[u] = u <= n;
    for (register int i = beg2[u]; i; i = nex[i])
        getsz(v[i]), sz[u] += sz[v[i]];
}

void DP(int u) {
    int pre = u <= n;
    for (register int i = beg2[u]; i; i = nex[i])
        DP(v[i]), Ans += (LL)pre * w[u] * sz[v[i]], pre += sz[v[i]];
    Ans += (LL)sz[u] * (all - sz[u]) * w[u];
}

int main()
{
#ifdef hany01
    File("loj2587");
#endif

    static int m, uu, vv;

    tot = n = read(), m = read();
    while (m --) uu = read(), vv = read(), add1(uu, vv), add1(vv, uu);

    For(i, 1, n) if (!dfn[i]) DFS(i, 0), getsz(i), all = sz[i], DP(i);
    printf("%lld\n", Ans << 1);

    return 0;
}