解题报告 之 HDU5325 Crazy Bobo
Description
Bobo has a tree,whose vertices are conveniently labeled by 1,2,...,n.Each node has a weight 
.
All the weights are distrinct.
A set with m nodes 
is
a Bobo Set if:
- The subgraph of his tree induced by this set is connected.
- After we sort these nodes in set by their weights in ascending order,we get 
,(that
is,
for
i from 1 to m-1).For any node 
in
the path from to 
(excluding and ),should
satisfy 
.
Your task is to find the maximum size of Bobo Set in a given tree.
Input
The input consists of several tests. For each tests:
The first line contains a integer n (
).
Then following a line contains n integers 
(
,all
the is
distrinct).Each of the following n-1 lines contain 2 integers 
and 
,denoting
an edge between vertices and ().
The sum of n is not bigger than 800000.
Output
For each test output one line contains a integer,denoting the maximum size of Bobo Set.
Sample Input
7 3 30 350 100 200 300 400 1 2 2 3 3 4 4 5 5 6 6 7
Sample Output
5
题目大意：给一个连通图，有n个节点，n-1条无向边并给出每个点的权值。让你找一个最大的子图满足子图中的连接权重相邻的两点 u,v 的道路中任意节点x的权值都比u、v的小。（其实好像看题的时候没看出来是这个意思，查了下才知道）。
分析：因为这个子图任意权重相邻的点都要满足这个条件，所以这个子图一定连通（否则一定存在权值相邻的点之间没有连通path）。然后注意到这个条件，这个子图任意权重相邻点之间的路要么是没有点，要么一定是呈现山谷状的，即两边最高，向中间递减。那么从图中的一个点开始，以它为最低点，向两边较高的扩展，然后再以同样的方式去搜索这两个较高点。那么任意两个权值相邻的点，往回找通路的时候一定都是呈现山谷状的。那么问题就转换成了，从某个点出发，去找到它能够扩展到的最大点数，深搜一次即可。
上代码：
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int MAXN = 5e6+10;
struct Edge
{
int to, nxt;
};
Edge edge[MAXN];
int head[MAXN];
int w[MAXN];
int num[MAXN];
int cnt, ans;
void addedge( int from, int to )
{
edge[cnt].to = to;
edge[cnt].nxt = head[from];
head[from] = cnt++;
}
int dfs( int u )
{
if(num[u]) return num[u];
num[u] = 1;
for(int i = head[u]; i != -1; i = edge[i].nxt)
{
int v = edge[i].to;
num[u]+=dfs( v );
}
return num[u];
}
int main()
{
int n;
while(scanf( "%d", &n ) == 1)
{
cnt = 0;
memset( num, 0, sizeof num );
memset( head, -1, sizeof head );
for(int i = 1; i <= n; i++)
scanf( "%d", &w[i] );
for(int i = 0; i < n - 1; i++)
{
int u, v;
scanf( "%d%d", &u, &v );
if(w[u] < w[v])
addedge( u, v );
else
addedge( v, u );
}
ans = 0;
for(int i = 1; i <= n; i++)
ans = max( ans, dfs( i ) );
cout << ans << endl;
}
return 0;
}
嗯就是这样！
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