9.18 Roads on a kingdom

题意

给定(n)个点(n)条边的连通图，要求删去一条边，使得剩下的图仍是一个连通图，并且图中距离最远的两个点的距离最小

解法

(n)个点(n)条边的连通图即为基环树

要删去一条边使得剩下的图仍是连通图，我们删除的一定是环上的边，剩下的图一定是一颗树

那么树上距离最远的两个点的距离就是树的直径

暴力枚举环上断边求直径是(O(N^2))的，考虑优化

单独考虑环上的点，把环上的点编号为(1)至(m)

断边后的直径有两种情况：

• 直径不经过环( ightarrow) 答案即为所有基环外向树中的直径

• 直径经过环 ( ightarrow) 此时又要分两种情况进行考虑

  • 使用了(1 o m)这条边

    预处理出倒(L)形路径的前缀后缀最大值

  • 未使用(1 o m)这条边

    直径一定是从某个子树的最深处出发，在环上遍历了某些点，再回到某个子树中去的

    预处理出(Pi)形路径的前缀后缀最大值

有很多细节，巨难打。。

代码

#include <bits/stdc++.h> 

using namespace std;

const int N = 1e6 + 10;

int read();

int n;

int cap;
int head[N], to[N], nxt[N], val[N];

int deg[N];

long long ans;

inline void add(int x, int y, int z) {
	to[++cap] = y, nxt[cap] = head[x], head[x] = cap, val[cap] = z; 
}

int p;
int cir[N], mrk[N], book[N];

long long DIS, len[N];

void Dfs_cir(int x) {
	book[x] = mrk[x] = 1, cir[++p] = x;
	for (int i = head[x]; i; i = nxt[i]) {
		if (!mrk[to[i]]) {
			Dfs_cir(to[i]);
			break;
		}
	}
}

long long f[N][5];

void Dfs_diametre(int x, int fa) {
	for (int i = head[x]; i; i = nxt[i]) {
		if (to[i] == fa || book[to[i]])	continue;
		Dfs_diametre(to[i], x);
		if (f[to[i]][1] + val[i] >= f[x][1])
			f[x][2] = f[x][1], f[x][1] = f[to[i]][1] + val[i];
		else
			f[x][2] = max(f[x][2], f[to[i]][1] + val[i]);
	}
}

int top;
int stk[N];

void Get_cir() {
	for (int i = 1; i <= n; ++i)
		if (deg[i] == 1)
			stk[++top] = i;
	while (top) {
		int u = stk[top--];
		mrk[u] = 1;
		for (int i = head[u]; i; i = nxt[i]) {
			deg[u]--, deg[to[i]]--;
			if (deg[to[i]] == 1)
				stk[++top] = to[i];
		}
	}
	for (int i = 1; i <= n; ++i)
		if (!mrk[i])	Dfs_cir(i);
}

void Get_diametre() {
	for (int i = 1; i <= p; ++i) {
		Dfs_diametre(cir[i], 0);
		ans = max(ans, f[cir[i]][1] + f[cir[i]][2]);
	}
}

long long pre[N], suf[N];
long long f1[N], f2[N], g1[N], g2[N];

void prepare_for_ans() {
	for (int i = 1; i < p; ++i)
		for (int j = head[cir[i]]; j; j = nxt[j]) 
			if (to[j] == cir[i + 1])  
				len[i] = val[j];
	
	for (int i = head[cir[p]]; i; i = nxt[i])
		if (to[i] == cir[1]) {
			DIS = val[i];
			break;	
		}
	
	for (int i = 1; i < p; ++i)			pre[i] = pre[i - 1] + len[i];
	for (int i = p - 1; i >= 1; --i)	suf[i] = suf[i + 1] + len[i];
	
	for (int i = 1; i <= p; ++i)   f1[i] = max(f1[i - 1], pre[i - 1] + f[cir[i]][1]);
	for (int i = p; i >= 1; --i)   f2[i] = max(f2[i + 1], suf[i] + f[cir[i]][1]);
	
	long long now = f[cir[1]][1];
	for (int i = 2; i <= p; ++i) {
		now += len[i - 1];
		g1[i] = max(g1[i - 1], now + f[cir[i]][1]);
		now = max(now, f[cir[i]][1]);
	}
	
	now = f[cir[p]][1];	
	for (int i = p - 1; i >= 1; --i) {
		now += len[i];
		g2[i] = max(g2[i + 1], now + f[cir[i]][1]);
		now	= max(now, f[cir[i]][1]);
	}
	
}

void Get_ans() {
	prepare_for_ans();
	
	long long res = g2[1];
	
	for (int i = 1; i < p; ++i) 
		res = min(res, max(max(g1[i], g2[i + 1]), f1[i] + f2[i + 1] + DIS));
	ans = max(ans, res);
}

int main() {
	
	n = read();
	
	for (int i = 1; i <= n; ++i) {
		int u = read(), v = read(), w = read();
		add(u, v, w), add(v, u, w);	
		deg[u]++, deg[v]++;
	}
	
	Get_cir();
	Get_diametre();
	Get_ans();
	
	printf("%lld
", ans);
	
	return 0;
}

#define gc getchar
int read() {
	int x = 0, c = gc();
	while (c < '0' || c > '9')	  c = gc();
	while (c >= '0' && c <= '9')  x = x * 10 + c - 48, c = gc();
	return x;	
}