51nod"省选"模测 A 树的双直径(树形dp)

题意

题目链接

Sol

比赛结束后才调出来。。不多说啥了，就是因为自己菜。

裸的up-down dp，维护一下一个点上下的直径就行，一开始还想了个假的思路写了半天。。

转移都在代码注释里

毒瘤题目卡空间

#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define LL long long
using namespace std;
const int MAXN = 6e5 + 10;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A> inline void debug(A a){cout << a << '\n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
void print(__int128 x) {
    if (x>9) print(x/10);
    putchar('0'+x%10);
}
int N, p[MAXN * 2], ve[MAXN * 2];
long long f[MAXN], g[MAXN], mxpre[MAXN], mxsuf[MAXN], dl[MAXN], ul[MAXN];
__int128 ans;
vector<Pair> v[MAXN];
void downdfs(int x, int fa) {
	f[x] = dl[x] = 0;
	for(auto &tp : v[x]) {
		int to = tp.fi, w = tp.se;
		if(to == fa) continue;
		downdfs(to, x);
		chmax(dl[x], f[x] + f[to] + w);
		chmax(f[x], f[to] + w);
		chmax(dl[x], dl[to]);
	}
}

void updfs(int x, int fa) {
	int cnt = 0;
	for(int i = 0; i < v[x].size(); i++)
		if(v[x][i].fi != fa) p[++cnt] = v[x][i].fi, ve[cnt] = v[x][i].se;

	mxpre[0] = 0; mxsuf[cnt + 1] = 0;
	for(int i = 1; i <= cnt; i++) mxpre[i] = max(mxpre[i - 1], f[p[i]] + ve[i]);
	for(int i = cnt; i >= 1; i--) mxsuf[i] = max(mxsuf[i + 1], f[p[i]] + ve[i]);
	//一个点向上的直径：
	//1: 从父亲的ul继承过来
	//2: 前后缀中的最大值f + 出边 + 入边
	//3: 父亲的g + 兄弟节点中最大的f + 出边
	//4: 前驱/后继 中的最大和次大
	//5: 前驱/后继 中的子树中的直径 

	for(int i = 1; i <= cnt; i++) {
		int to = p[i], w = ve[i];
		chmax(g[to], g[x] + w);
		chmax(g[to], max(mxpre[i - 1], mxsuf[i + 1]) + w);
		chmax(ul[to], ul[x]);//1
		chmax(ul[to], mxpre[i - 1] + mxsuf[i + 1]);//2
		chmax(ul[to], g[x] + max(mxpre[i - 1], mxsuf[i + 1]));//3
	}

	long long mx1 = -1e18, mx2 = -1e18, mx3 = -1e18;
	for(int i = 1; i <= cnt; i++) {
		chmax(ul[p[i]], max(mx1 + mx2, mx3)); int tmp = f[p[i]] + ve[i];
		if(tmp > mx1) chmax(mx2, mx1), mx1 = tmp;
		else if(tmp > mx2) mx2 = tmp;
		chmax(mx3, dl[p[i]]);
	}
	mx1 = -1e18, mx2 = -1e18, mx3 = -1e18;
	for(int i = cnt; i >= 1; i--) {
		chmax(ul[p[i]], max(mx1 + mx2, mx3)); int tmp = f[p[i]] + ve[i];
		if(tmp > mx1) chmax(mx2, mx1), mx1 = tmp;
		else if(tmp > mx2) mx2 = tmp;
		chmax(mx3, dl[p[i]]);
	}

	for(int i = 0; i < v[x].size(); i++)
		if(v[x][i].fi != fa) updfs(v[x][i].fi, x);

}
signed main() {
	//freopen("a.in", "r", stdin);
    N = read();
    for(int i = 1; i <= N - 1; i++) {
    	int x = read(), y = read(), w = read();
    	v[x].push_back(MP(y, w));
    	v[y].push_back(MP(x, w));
	}
    downdfs(1, 0);
    updfs(1, 0);
    for(int i = 2; i <= N; i++) chmax(ans, (__int128) dl[i] * ul[i]);

    for(int i = 1; i <= N; i++) for(auto &tp : v[i]) tp.se = -tp.se;
    memset(ul, 0, sizeof(ul));
    memset(g, 0, sizeof(g));
    downdfs(1, 0);
    updfs(1, 0);
    for(int i = 2; i <= N; i++) chmax(ans, (__int128) dl[i] * ul[i]);
   	print(ans);
   	//cout << ans;
    return 0;
}