2015 多校联赛 ——HDU5325(DFS)
2015-07-30 10:39
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Crazy Bobo
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 1215 Accepted Submission(s): 366
Problem Description
Bobo has a tree,whose vertices are conveniently labeled by 1,2,...,n.Each node has a weight wi.
All the weights are distrinct.
A set with m nodes v1,v2,...,vm 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 u1,u2,...,um,(that
is,wui<wui+1 for
i from 1 to m-1).For any node x in
the path from ui to ui+1(excluding ui and ui+1),should
satisfy wx<wui.
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 (1≤n≤500000).
Then following a line contains n integers w1,w2,...,wn (1≤wi≤109,all
the wi is
distrinct).Each of the following n-1 lines contain 2 integers ai and bi,denoting
an edge between vertices ai and bi (1≤ai,bi≤n).
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
Source
2015 Multi-University Training Contest 3
题意:给一个n,然后给n个点的值,再输入n-1条边,构成一个树(当时没看懂要求啥 QAQ)后来看别人的解题报告大致明白,
3 30 350 100 200 300 400 可以建成以下有向图
1 -->2-->3<--4-->5-->6-->7 当在点4的时候,能总共走过5个点,所以输出5(感觉不难 - -!! 论英语的重要性)
用深搜要手动扩栈,C++提交,否则会出现 Runtime Error (ACCESS_VIOLATION) //表示新手并不知道这是啥
当时就是因为提交出了这个,以为这题自己想得太简单,就放弃了 - -
而且官方给的测试数据,正确代码也只能输出一半,搞得一直以为自己错了!!
#include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <algorithm> #pragma comment(linker, "/STACK:102400000,102400000")//手动扩栈 typedef long long ll; using namespace std; const int maxn= 5e5 + 5; const ll INF = 1000000000000000000; int p[maxn]; vector<int>q[maxn]; int num[maxn]; void dfs(int u) { num[u]++; int len = q[u].size(); for(int i = 0;i < len;i++) { int son = q[u][i]; if(!num[son]) dfs(son); num[u]+=num[son]; } } int main() { int n; //freopen("10.txt","r",stdin); while(scanf("%d",&n) != EOF) { for(int i = 1;i <= n;i++) scanf("%d",&p[i]); int a,b; for(int i = 1;i <= n;i++) q[i].clear(); for(int i = 1;i < n;i++) { scanf("%d%d",&a,&b); if(p[a] < p[b]) q[a].push_back(b); else q[b].push_back(a); } int ans = 0; memset(num,0,sizeof(num)); for(int i = 1;i <= n;i++) { if(!num[i]) dfs(i); ans= max(ans,num[i]); } printf("%d\n",ans); } }
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