POJ 3107 - Godfather 树形DP..vector慎用...

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提交超时..实在觉得没什么好优化的...最多改回至底而上的BFS..但好麻烦,记一堆东西..看discuss才知道主要是vector的原因..改成手写链表..500MS过,,,

选择任意一个点做树的树的root...统计每个点的子树元素个数情况..对于不是root的点..将所有点数N减去当前子树的元素个数num.作为该点的另一个孩子...

Program:

#include<iostream>#include<stdio.h>#include<string.h>#include<set>#include<ctime>#include<algorithm>#include<queue>#include<cmath>#include<map>#define oo 1000000007#define ll long long#define pi acos(-1.0)#define MAXN 50005using namespace std;  struct node{      int x,y,next;}line[MAXN*2];  int n,AnsNum,AnsData,ans[MAXN],_next[MAXN];bool used[MAXN];void addline(int x,int y,int m){      line[m].next=_next[x],_next[x]=m;      line[m].x=x,line[m].y=y;      return;}int dfs(int x){      int MaxSub=0,num=0,t,k;      k=_next[x];      while (k)       {           if (!used[line[k].y])           {                 used[line[k].y]=true;                 t=dfs(line[k].y);                 MaxSub=max(t,MaxSub);                 num+=t;                 used[line[k].y]=false;           }           k=line[k].next;      }      MaxSub=max(MaxSub,n-(num+1));      if (MaxSub==AnsData) ans[++AnsNum]=x;      else       if (MaxSub<AnsData)      {              AnsData=MaxSub;             AnsNum=0,ans[++AnsNum]=x;      }      return num+1;}int main(){        int i,num;       while (~scanf("%d",&n))      {              memset(_next,0,sizeof(_next));              for (i=1;i<n;i++)              {                     int x,y;                     scanf("%d%d",&x,&y);                     addline(x,y,i*2-1);                     addline(y,x,i*2);               }              memset(used,false,sizeof(used));              AnsData=oo; used[1]=true;              dfs(1);              sort(ans+1,ans+1+AnsNum);              printf("%d",ans[1]);              for (i=2;i<=AnsNum;i++) printf(" %d",ans[i]);              printf("\n");      }      return 0;}