题解 【luogu P2680 NOIp提高组2015 运输计划】

题目链接

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题解
题意
一棵树上有(m)条路径，可以将其中一条边的权值改为0，问最长的路径最短是多少
分析

• 最短的路径最长自然想到二分最长路径，设其为(dis)
• 关键在于如何check
• check的关键又是将哪条边改为0
• 贪心，如果所有超过(dis)的路径能在一条边上重合，则将那条边改为0，之后再判断改为0后是否最大的路径小于(dis)；若无法将所有超过(dis)的边重合在一条边上，直接return false;

做法

• 算两个点之间的路径长用dfs + LCA来实现
• 判断路径之间的重合用树上差分来实现
• 这里用的是倍增

注意事项

无向边要把数组开两倍！！！

代码

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>

using namespace std;

const int MAXN = 500500;
int n, m;
int logn;
int u[MAXN], v[MAXN], lca[MAXN];
int vis[MAXN], dep[MAXN], fa[MAXN][25];
int dfn[MAXN], Index, Edge_cnt;
int Max_dis = -1, treeC[MAXN], predis[MAXN], dissum[MAXN], distence[MAXN];

int ecnt;
struct node
{
    int v;
    int w;
    node *next;
}pool[MAXN << 1], *head[MAXN << 1];

void addedge(int u, int v, int w)
{
    node *p = &pool[++ecnt], *q = &pool[++ecnt];
    p->v = v, p->w = w, p->next = head[u], head[u] = p;
    q->v = u, q->w = w, q->next = head[v], head[v] = q;
}

void dfs(int u)
{
    int v;
    dfn[++Index] = u;
    vis[u] = 1;
    for(node *p = head[u]; p; p = p->next)
        if(!vis[v = p->v])
        {
        	dep[v] = dep[u] + 1;
        	dissum[v] = dissum[u] + p->w;
        	fa[v][0] = u;
        	predis[v] = p->w;
        	dfs(v);
        }	      
}

int LCA(int u, int v)
{
    if(dep[u] < dep[v]) swap(u, v);
    for(int i = 20; i >= 0; i--)
        if(fa[u][i] != 0 && dep[fa[u][i]] >= dep[v])
            u = fa[u][i];
    if(u == v) return u;
    for(int i = 20; i >= 0; i--)
        if(fa[u][i] != fa[v][i])
            u = fa[u][i], v = fa[v][i];
    return fa[u][0];
}

bool check(int dis)
{
    memset(treeC, 0, sizeof(treeC));
    Edge_cnt = 0;
    for(int i = 1; i <= m; i++)
        if(distence[i] > dis)
        {
        	++Edge_cnt;
        	treeC[u[i]]++;
        	treeC[v[i]]++;
        	treeC[lca[i]] -= 2;
        }
    for(int i = n; i >= 1; i--)
    {
    	int t = dfn[i];
    	treeC[fa[t][0]] += treeC[t];
    	if(dis >= Max_dis - predis[t] && treeC[t] == Edge_cnt)
    	    return true;
    }
    return false;
}

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n - 1; i++)
    {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        addedge(u, v, w);
    }
    dep[1] = 1, dep[0] = -1;
    dfs(1);
	for(int j = 1; j <= 20; j++)
        for(int i = 1; i <= n; i++)
            fa[i][j] = fa[fa[i][j - 1]][j - 1];
    for(int i = 1; i <= m; i++) 
    {
        scanf("%d%d", &u[i], &v[i]);
        lca[i] = LCA(u[i], v[i]);
        distence[i] = dissum[u[i]] + dissum[v[i]] - dissum[lca[i]] * 2;
        Max_dis = max(Max_dis, distence[i]);
    }
    int ans = -1;
    int l = 0, r = Max_dis;
    while(l <= r)
    {
        int mid = (l + r) / 2;
        if(check(mid)) r = mid - 1, ans = mid;
        else l = mid + 1;
    }
    printf("%d
", ans);
    return 0;
}