hdu 5325 Crazy Bobo 多校1010

Crazy Bobo

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 670    Accepted Submission(s): 208

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

73 30 350 100 200 300 4001 22 33 44 55 66 7

 

Sample Output

5

 

Source

2015 Multi-University Training Contest 3

这里给出两种方法。【其实差不多。。

构建出一个单向图，每次dfs出一个没有更新的点去加上能更新的点。用一个数组记录每个点的值，也在这里更新。最后输出最大的

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <stdio.h>#include <queue>#include <vector>#include <string.h>using namespace std;#define Max 500010vector<int> edge[Max];int w[Max],ans[Max],n;void dfs(int a){    ans[a]=1;    int len=edge[a].size();    for(int i=0;i<len;i++)    {        int v=edge[a][i];        if(!ans[v]) dfs(v);        ans[a]+=ans[v];    }}int main(){    while(~scanf("%d",&n))    {        memset(ans,0,sizeof(ans));        for(int i=1;i<=n;i++)        {            scanf("%d",&w[i]);            edge[i].clear();        }        int a,b;        for(int i=1;i<n;i++)        {            scanf("%d%d",&a,&b);            if(w[a]<w[b]) edge[a].push_back(b);            else edge[b].push_back(a);        }        for(int i=1;i<=n;i++)        {            if(ans[i]) continue;            dfs(i);        }        int maxx=-1;        for(int i=1;i<=n;i++)        {            maxx=max(maxx,ans[i]);        }        cout<<maxx<<endl;    }    return 0;}

第一句话是hdu防爆栈的。。。

先建图，按值排序 

 然后
for i = 1 to n
f[a[i].w] = Σf[v] + 1;(v为与a[i].i邻接的点)

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <string.h>#include <math.h>#include <queue>#include <stack>#include <stack>#include <set>#include <map>#include <fstream>using namespace std;#define Max 500010int vis[Max],v[Max],k[Max],top;vector<int>mp[Max];int w[Max];struct pt{    int w,p;};pt r[Max];struct ppt{    int w;};ppt qq[Max];bool cmp(pt a,pt b){    return a.w>b.w;}int main(){    int n;    while(cin>>n)    {        int a,b;        memset(k,0,sizeof(k));        for(int i=1;i<=n;i++)        {            scanf("%d",&r[i].w);            //cin>>r[i].w;            r[i].p=i;            mp[i].clear();        }        sort(r+1,r+n+1,cmp);        for(int i=1;i<n;i++)        {            scanf("%d%d",&a,&b);            //cin>>a>>b;            mp[a].push_back(b);            mp[b].push_back(a);        }        int res=0;        for(int i=1;i<=n;i++)        {            int u=r[i].p;            int len=mp[u].size();            k[u]++;            for(int i=0;i<len;i++)            {                int v=mp[u][i];                k[u]+=k[v];            }        }        res=-1;        for(int i=1;i<=n;i++) {res=max(k[i],res);}        cout<<res<<endl;        //outfile.close();    }    return 0;}