【USACO Contest】Holiday 2010 Bonus Competition, Problem 1: Cow Politics (cowpol)

还没看完整个题目，我就想到：啊，树的最长路！

于是我就轻率地写了个树的最长路的算法，严重TLE。因为每求一次的时间是O(n)，k次就是O(n*k)，要是n跟k都大（一开始忘了这一点），就成了跟o(n^2)差不多了。跟最笨的算法——从每个节点出发做一次DFS是一个等级的。

 

附：第一版TLE代码：

儿子兄弟表示法+改进树的最长路

/*ID: gaojiaw1PROG: cowpolLANG: C++*/#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <fstream>#include <string>using namespace std;int a[200001],p[200001],f[200001],lc[200001],rc[200001],pt,ans;int w(int x){ if(f[x]!=-1)return f[x]; int i,t,b=(a[x]==pt),max=0,max2=0; for(i=lc[x];i!=0;i=rc[i]) { t=w(i); if(t>max) { max2=max; max=t; } else if(t>max2) max2=t; } if(max) { f[x]=max+1; if(max2) { if(max+max2>ans)ans=max+max2; } else if(b) { if(max>ans)ans=max; } } else f[x]=b; return f[x];}int main() { ofstream fout ("cowpol.out"); ifstream fin ("cowpol.in"); int n,k,i,rt; fin>>n>>k; for(i=1;i<=n;i++) { fin>>a[i]>>p[i]; if(!p[i])rt=i; else { if(lc[p[i]]==0) lc[p[i]]=i; else { rc[i]=rc[lc[p[i]]]; rc[lc[p[i]]]=i; } } } for(pt=1;pt<=k;pt++) { memset(f,0xff,sizeof(f)); ans=0; w(rt); fout<<ans<<endl; } return 0;}

 

于是我就看了Analysis。

大概是这么说的：

1.先找到每个党在树中最深的节点。它一定是最长的路的一端。（orz这个想法，太巧了，有点像某年noip的树网的核一样巧……）

2.用tarjan算法求LCA。这里的LCA就是这个点与该党最深的节点间的最近公共祖先。

在tarjan(x)中，如果x是该党最深节点，则用它来更新所有该党的节点的LCA；

如果x不是该党最深节点，就用该党最深节点来更新它。

3.x,y距离=depth(x)+depth(y)-2*(LCA(x,y))

 

优化后代码如下：

/*ID: gaojiaw1PROG: cowpolLANG: C++*/#include <iostream>#include <fstream>using namespace std;int a[200001], //party p[200001], //parent f[200001], //LCA lc[200001], //left child rc[200001], //next sibling d[200001], //depth v[200001], //visited node far[200001], //the deepest node of one party anc[200001], //ancestor calculated in lca function set[200001], //set of a node n,k;//calculate depths for each nodevoid dfs(int x,int dep){ //cout<<x<<' '<<dep<<endl; d[x]=dep; if(dep>d[far[a[x]]])far[a[x]]=x; for(int i=lc[x];i!=0;i=rc[i])dfs(i,dep+1);}//find setint find(int x){ if(set[x]==x)return x; else set[x]=find(set[x]); return set[x];}//union two setsvoid join(int x,int y){ set[find(x)]=find(y);}//Tarjan LCA algorithmvoid tarjan(int x){ int i; set[x]=anc[x]=x; for(i=lc[x];i!=0;i=rc[i]) { tarjan(i); join(x,i); anc[find(x)]=x; } v[x]=1; /* x is the deepest node of its party: * set f[] of every other node * with the LCA of that node and x */ if(x==far[a[x]]) { for(i=1;i<=n;i++) if(v[i]&&a[i]==a[x]&&i!=x) f[i]=anc[find(i)]; } /* x is not the deepest node of its party: * set f[x] with the LCA of * x and the deepest node of its party */ else { i=far[a[x]]; if(v[i])f[x]=anc[find(i)]; }}int main(){ ofstream fout ("cowpol.12gjw1.out"); ifstream fin ("cowpol.12.in"); int i,j,rt,max,t; fin>>n>>k; for(i=1;i<=n;i++) { fin>>a[i]>>p[i]; if(!p[i])rt=i; //i is root else { if(lc[p[i]]==0)lc[p[i]]=i; //make i left child of p[i] else { rc[i]=rc[lc[p[i]]]; //make i next sibing of lc[p[i]] rc[lc[p[i]]]=i; } } far[a[i]]=i; } dfs(rt,0); tarjan(rt); for(j=1;j<=k;j++) { t=far[j]; max=0; for(i=1;i<=n;i++) if(a[i]==j && i!=t && d[i]+d[t]-2*d[f[i]]>max) max=d[i]+d[t]-2*d[f[i]]; fout<<max<<endl; } return 0;}

 

结果还是TLE。

 

我发现我犯傻了……枚举每个党的成员的时候不应该从1到n枚举然后判断，而是应该先存到一个邻接表里面。（用vector实现）

 

第三版：

/*ID: gaojiaw1PROG: cowpolLANG: C++*/#include <fstream>#include <iostream>#include <vector>using namespace std;int a[200001], //party p[200001], //parent f[200001], //LCA lc[200001], //left child rc[200001], //next sibling d[200001], //depth v[200001], //visited node far[100001], //the deepest node of one party anc[200001], //ancestor calculated in lca function set[200001], //set of a node n,k;vector<int> party[100001];//calculate depths for each nodevoid dfs(int x,int dep){ d[x]=dep; if(dep>d[far[a[x]]])far[a[x]]=x; for(int i=lc[x];i!=0;i=rc[i])dfs(i,dep+1);}//find setint find(int x){ if(set[x]==x)return x; else set[x]=find(set[x]); return set[x];}//union two setsvoid join(int x,int y){ set[find(x)]=find(y);}//Tarjan LCA algorithmvoid tarjan(int x){ unsigned int i,y; set[x]=anc[x]=x; for(i=lc[x];i!=0;i=rc[i]) { tarjan(i); join(x,i); anc[find(x)]=x; } v[x]=1; /* x is the deepest node of its party: * set f[] of every other node * with the LCA of that node and x */ if(x==far[a[x]]) { for(i=0;i<party[a[x]].size();i++) { y=party[a[x]][i]; if(v[y]&&y!=x) f[y]=anc[find(y)]; } } /* x is not the deepest node of its party: * set f[x] with the LCA of * x and the deepest node of its party */ else { i=far[a[x]]; if(v[i])f[x]=anc[find(i)]; }}int main(){ ofstream fout ("cowpol.out"); ifstream fin ("cowpol.in"); unsigned int i,j,rt,max,t; fin>>n>>k; for(i=1;i<=n;i++) { fin>>a[i]>>p[i]; if(!p[i])rt=i; //i is root else { if(lc[p[i]]==0)lc[p[i]]=i; //make i left child of p[i] else { rc[i]=rc[lc[p[i]]]; //make i next sibing of lc[p[i]] rc[lc[p[i]]]=i; } } far[a[i]]=i; party[a[i]].push_back(i); } dfs(rt,0); tarjan(rt); for(j=1;j<=k;j++) { t=far[j]; max=0; for(i=0;i<party[j].size();i++) { int u=party[j][i]; if(d[u]+d[t]-2*d[f[u]]>max && u!=t) max=d[u]+d[t]-2*d[f[u]]; } fout<<max<<endl; } return 0;}

 

但是不知道为什么，交上去在第12、13个测试数据会有segmentation fault，而在我的机器上跑的好好的，答案也对，而且我开的空间远远没有MLE。我为了弄清问题，从windows换到了linux，也run的好好的。莫非usaco的机器问题？呃……罢了罢了……就当我Ac了算了-_-!

 

  > Run 1: OK [0.000 secs, 11608 KB]
  > Run 2: OK [0.000 secs, 11608 KB]
  > Run 3: OK [0.000 secs, 11608 KB]
  > Run 4: OK [0.011 secs, 11608 KB]
  > Run 5: OK [0.022 secs, 11636 KB]
  > Run 6: OK [0.086 secs, 11608 KB]
  > Run 7: OK [0.335 secs, 11872 KB]
  > Run 8: OK [0.151 secs, 17656 KB]  > Run 9: OK [1.112 secs, 19200 KB]
  > Run 10: OK [0.259 secs, 11608 KB]
  > Run 11: OK [0.346 secs, 12812 KB]

  > Run 12: Execution error: Your program had this runtime error:
    Exceeded memory or invalid memory reference. The program ran for
    0.302 CPU seconds before the error. It used 12796 KB of memory.

> Run 13: Execution error: Your program (`cowpol') exited with    signal #11 (segmentation violation [maybe caused by accessing
    memory out of bounds, array indexing out of bounds, using a bad
    pointer (failed open(), failed malloc), or going over the maximum
    specified memory limit]). The program ran for 0.292 CPU seconds
    before the signal. It used 12664 KB of memory.