hdu 5325 Crazy Bobo (树形dp)

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Crazy Bobo

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1077 Accepted Submission(s): 326

Problem Description

Bobo has a tree,whose vertices are conveniently labeled by 1,2,...,n.Each node has a weight

Input

The input consists of several tests. For each tests:
The first line contains a integer n (

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

YY一下，想到只要保证一棵子树的节点到其所有的叶子节点的路径都是满足递增的，找出一棵这样的最大的子树即可。

先从下往上dp一遍，然后再从上往下，把父亲的信息传送下来，嗯，然后就以O(n)得到了所需要的结果

不过直接dfs会爆栈，所以得开一下栈，然后用C++提交

  1 //#####################
  2 //Author:fraud
  3 //Blog: http://www.cnblogs.com/fraud/
  4 //#####################
  5 #pragma comment(linker, "/STACK:102400000,102400000")
  6 #include <iostream>
  7 #include <sstream>
  8 #include <ios>
  9 #include <iomanip>
 10 #include <functional>
 11 #include <algorithm>
 12 #include <vector>
 13 #include <string>
 14 #include <list>
 15 #include <queue>
 16 #include <deque>
 17 #include <stack>
 18 #include <set>
 19 #include <map>
 20 #include <cstdio>
 21 #include <cstdlib>
 22 #include <cmath>
 23 #include <cstring>
 24 #include <climits>
 25 #include <cctype>
 26 using namespace std;
 27 #define XINF INT_MAX
 28 #define INF 0x3FFFFFFF
 29 #define MP(X,Y) make_pair(X,Y)
 30 #define PB(X) push_back(X)
 31 #define REP(X,N) for(int X=0;X<N;X++)
 32 #define REP2(X,L,R) for(int X=L;X<=R;X++)
 33 #define DEP(X,R,L) for(int X=R;X>=L;X--)
 34 #define CLR(A,X) memset(A,X,sizeof(A))
 35 #define IT iterator
 36 typedef long long ll;
 37 typedef pair<int,int> PII;
 38 typedef vector<PII> VII;
 39 typedef vector<int> VI;
 40 int Scan() {
 41     int res=0, ch;
 42     while(ch=getchar(), ch<'0'||ch>'9');
 43     res=ch-'0';
 44     while((ch=getchar())>='0'&&ch<='9')
 45         res=res*10+ch-'0';
 46     return res;
 47 }
 48 void Out(int a) {
 49     if(a>9)
 50         Out(a/10);
 51     putchar(a%10+'0');
 52 }
 53 int w[500010];
 54 vector<int>G[500010];
 55 int dp[500010];
 56 void dfs1(int u,int fa){
 57     dp[u] = 1;
 58     REP(i,G[u].size()){
 59         int v = G[u][i];
 60         if(v == fa)continue;
 61         dfs1(v,u);
 62         if(w[v] > w[u])dp[u] += dp[v];
 63     }
 64 }
 65 void dfs2(int u,int fa){
 66     if(fa!=-1){
 67         if(w[fa] > w[u])dp[u] += dp[fa];
 68     }
 69     REP(i,G[u].size()){
 70         int v = G[u][i];
 71         if(v == fa)continue;
 72         dfs2(v,u);
 73     }
 74 }
 75 int main()
 76 {
 77     //ios::sync_with_stdio(false);
 78     int n;
 79     while(scanf("%d",&n)!=EOF){
 80         REP(i,n)w[i] = Scan();
 81         REP(i,n)G[i].clear();
 82         int u,v;
 83         REP(i,n-1){
 84             u = Scan();
 85             v = Scan();
 86             u--;v--;
 87             G[u].PB(v);
 88             G[v].PB(u);
 89         }
 90         dfs1(0,-1);
 91         dfs2(0,-1);
 92         int ans = 0;
 93         REP(i,n)ans = max(dp[i],ans);
 94         Out(ans);
 95         puts("");
 96     }
 97     return 0;
 98 }
 99 
100 
101 /*
102 7
103 5 1 7 2 3 8 6
104 1 2
105 1 3
106 2 4
107 2 5
108 3 6
109 3 7
110 9
111 1 2 3 4 5 6 7 8 9
112 9 1
113 1 2
114 4 2
115 2 3
116 3 7
117 3 6
118 2 5
119 5 8
120 9
121 1 2 3 4 5 6 7 8 9
122 1 9
123 9 8
124 1 5
125 1 2
126 2 7
127 7 4
128 2 6
129 6 3
130 */

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原文地址：https://www.cnblogs.com/fraud/p/4687007.html