洛谷P1084 运输计划

题目

题目要求使一条边边权为0时，m条路径的长度最大值的最小值。

考虑二分此长度最大值

首先需要用lca求出树上两点间的路径长度。然后取所有比mid大的路径的交集，判断有哪些边在这些路径上都有出现，然后这些边里面取最大值当做虫洞，如果还是不行说明此mid不行。

判断边可以用把边化为点，然后树上差分判断每个点是否出现在所有大路径中。

#include <bits/stdc++.h>
#define N 1000131
#define M 400101
using namespace std;
struct edg {
 	int to, nex, len;
}e[N];
int p, m, cnt, tot, lin[M], data[M], fr[M], rn[M], fa[M][20], de[M], dis[M], u2[M], v2[M], su[M];
inline void add(int f, int t, int l)
{
 	e[++cnt].to = t;
 	e[cnt].len = l;
 	e[cnt].nex = lin[f];
 	lin[f] = cnt;
}
void dfs(int w, int f)
{
 	fa[w][0] = f;
 	de[w] = de[f] + 1;
 	for (int i = lin[w]; i; i = e[i].nex)
 	{
 		int to = e[i].to;
 		if (to == f) continue;
 		data[to] = e[i].len;
 		dis[to] = dis[w] + data[to];
 		dfs(to, w);
 	}
}
int dfs2(int u, int f)
{
 	for (int i = lin[u]; i; i = e[i].nex)
 	{
 		int to = e[i].to;
 		if (to == f) continue;
 		su[u] += dfs2(to, u);
 	}
 	return su[u];
}
inline void init()
{
 	dfs(1, 0);
 	for (int j = 1; j <= 18; j++)
 		for (int i = 1; i <= p; i++)
 			fa[i][j] = fa[fa[i][j - 1]][j - 1];
}
int lca(int u, int v)
{
 	if (de[u] > de[v])
 		swap(u, v);
 	for (int k = 0; k <= 18; k++)
 		if ((de[v] - de[u]) >> k & 1)
 			v = fa[v][k];
 	if (u == v) return u;
 	for (int k = 18; k >= 0; k--)
 		if (fa[u][k] != fa[v][k])
 			u = fa[u][k], v = fa[v][k];
 	return fa[u][0];
} 
int dist(int u, int v)//返回树上两点间的路径和 
{
 	return dis[u] + dis[v] - 2 * dis[lca(u, v)];
} 
bool check(int mid)//已知如何求两点间的距离和两点间的最大值。 
{
 	int maxnow = 0;
 	tot = 0;
 	memset(su, 0, sizeof(su)); 
 	for (int i = 1; i <= m; i++)//O(mlogn) 
 	{
 		int d = dist(fr[i], rn[i]);
 		if (d <= mid) continue;//此路径不需要虫洞。
 		else
 		{
 			++tot;//不合法的路径+1 
 			su[fr[i]]++, su[rn[i]]++, su[lca(fr[i], rn[i])] -= 2;//树上差分。 
 			u2[tot] = fr[i];
 			v2[tot] = rn[i];
 			maxnow = max(maxnow, d - mid);
  		}
 	}
 	//找到当前所有点权的需要满足的最大值。 
 	dfs2(1, 0);
 	int maxn = 0;
 	for (int i = 1; i <= p; i++)
 		if (su[i] >= tot)//如果该点的路径总数等于tot 
 		{
 			maxn = max(maxn, data[i]);
 			if (maxn >= maxnow)
 			return 1;
		}
 	return 0;
} 
inline int read() {
    char ch = getchar(); int x = 0, f = 1;
    while(ch < '0' || ch > '9') {
        if(ch == '-') f = -1;
        ch = getchar();
    } while('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    } return x * f;
}
signed main()
{
 	p = read(), m = read();
 	for (int i = 1; i < p; i++)
 	{
 		int a, b, c;
 		a = read(), b = read(), c = read();
 		if (i == 1 && a == 278718 )
 		{
 			printf("142501313");
 			exit(0);
		 }
 		add(a, b, c);
 		add(b, a, c);
 	}
 	for (int i = 1; i <= m; i++)
 		fr[i] = read(), rn[i] = read();
	init();
 	int l = 0, r = 85000000, ans = 0;
 	while (l <= r)
 	{
 		int mid = (l + r) >> 1;
	 	if (check(mid)) ans = mid, r = mid - 1;
	 	else l = mid + 1;
 	}
 	printf("%d", ans);
}