[ZJOI2012]旅游

看懂题意就好了。

就是相邻的城市连边，然后求图的直径。

/**

 * Problem:Journey

 * Author:Shun Yao

 * Time:2013.6.3

 * Result:Accepted

 */



#include <cstring>

#include <cstdio>

#include <algorithm>



const long Maxn = 200005;



long abs(long x) {

	return x < 0 ? -x : x;

}

long max(long x, long y) {

	return x > y ? x : y;

}

long min(long x, long y) {

	return x < y ? x : y;

}



class line {

public:

	long u, v, i;

	line() {}

	~line() {}

	line(long U, long V, long I):u(min(U, V)), v(max(U, V)), i(I) {}

} l[Maxn << 1];



bool cmpl(line a, line b) {

	return a.u != b.u ? a.u < b.u : a.v < b.v;

}



class gnode {

public:

	long v;

	gnode *next;

	gnode() {}

	~gnode() {}

	gnode(long V, gnode *ne):v(V), next(ne) {}

} *g[Maxn];



void add(long x, long y) {

	g[x] = new gnode(y, g[x]);

	g[y] = new gnode(x, g[y]);

}



long n, tot;

long q[Maxn], *L, *R;

long f[Maxn], d[Maxn][2];

long ans;

int main() {

	static long i, a, b, c;

	static gnode *e;

	freopen("journey.in", "r", stdin);

	freopen("journey.out", "w", stdout);

	scanf("%ld", &n);

	tot = 0;

	for (i = 1; i <= n - 2; ++i) {

		scanf("%ld%ld%ld", &a, &b, &c);

		if (abs(a % n - b % n) > 1 && abs(a - b) > 1)

			l[++tot] = line(a, b, i);

		if (abs(a % n - c % n) > 1 && abs(a - c) > 1)

			l[++tot] = line(a, c, i);

		if (abs(b % n - c % n) > 1 && abs(b - c) > 1)

			l[++tot] = line(b, c, i);

	}

	std::sort(l + 1, l + tot + 1, cmpl);

	for (i = 1; i <= tot; i += 2)

		add(l[i].i, l[i + 1].i);

	memset(f, 0, sizeof f);

	R = (L = &(*q = 1)) + 1;

	f[1] = -1;

	while (L != R) {

		for (e = g[*L]; e; e = e->next)

			if (!f[e->v])

				f[*(R++) = e->v] = *L;

		++L;

	}

	ans = 0;

	do {

		--L;

		d[*L][0] = 0;

		d[*L][1] = 0;

		for (e = g[*L]; e; e = e->next)

			if (f[e->v] == *L) {

				if (d[*L][0] < d[e->v][0] + 1) {

					d[*L][1] = d[*L][0];

					d[*L][0] = d[e->v][0] + 1;

				} else if (d[*L][1] < d[e->v][0] + 1)

					d[*L][1] = d[e->v][0] + 1;

			}

		ans = max(ans, d[*L][0] + d[*L][1] + 1);

	} while (L != q);

	printf("%ld", ans);

	fclose(stdin);

	fclose(stdout);

	return 0;

}