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ClosestCommon Ancestors·POJ1470

2024-07-15 10:07:42   219人阅读

传送门：http://poj.org/problem?id=1470

 

Closest Common Ancestors
Time Limit: 2000MS Memory Limit: 10000K
   

Description

Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines the closest common ancestor of u and v in the tree. The closest common ancestor of two nodes u and v is the node w that is an ancestor of both u and v and has the greatest depth in the tree. A node can be its own ancestor (for example in Figure 1 the ancestors of node 2 are 2 and 5)

Input

The data set, which is read from a the std input, starts with the tree description, in the form:

nr_of_vertices
vertex:(nr_of_successors) successor1 successor2 ... successorn
...
where vertices are represented as integers from 1 to n ( n <= 900 ). The tree description is followed by a list of pairs of vertices, in the form:
nr_of_pairs
(u v) (x y) ...

The input file contents several data sets (at least one).
Note that white-spaces (tabs, spaces and line breaks) can be used freely in the input.

Output

For each common ancestor the program prints the ancestor and the number of pair for which it is an ancestor. The results are printed on the standard output on separate lines, in to the ascending order of the vertices, in the format: ancestor:times
For example, for the following tree:

Sample Input

55:(3) 1 4 21:(0)4:(0)2:(1) 33:(0)6(1 5) (1 4) (4 2)      (2 3)(1 3) (4 3)

Sample Output

2:15:5

Hint

Huge input, scanf is recommended.

Source

Southeastern Europe 2000
 
第一个Tarjan LCA。。WA了N次的原因居然是。。。。。。。。。。。。。。。。。。。。。。输出没带 “  ：” 。。。视力拙计
 
Codes:
 1 #include<set> 2 #include<queue> 3 #include<vector> 4 #include<cstdio> 5 #include<cstring> 6 #include<iostream> 7 #include<algorithm> 8 using namespace std; 9 const int N = 910;10 #define For(i,n) for(int i=1;i<=n;i++)11 #define Rep(i,l,r) for(int i=l;i<=r;i++)12 13 struct edge{14     int t,next;15 }E[N*2];16 int head[N];17 vector<int> Q[N];18 int indeg[N],fa[N],anc[N],Ans[N],Hash[N][N];19 int Es,n,m,x,y,tot,root;char ch;20 bool vis[N];21 void makelist(int s,int t){22     E[Es].t = t; E[Es].next = head[s];23     head[s] = Es++;24 }25 26 int read(){27     ch = getchar();int num = 0;28     while(ch>‘9‘||ch<‘0‘) ch = getchar();29     while(ch>=‘0‘&&ch<=‘9‘){30         num = num * 10 + ch - ‘0‘;31         ch = getchar();32     }33     return num;34 }35 36 int find(int i){37     return (fa[i]==i)?(i):(find(fa[i]));38 }39 40 void LCA(int i){41     anc[i] = i;42     for(int p = head[i];p!=-1;p=E[p].next){43         if(vis[E[p].t]) continue;44         LCA(E[p].t);45         fa[find(E[p].t)] = find(i);46         anc[find(i)] = i;47     }48     vis[i] = true;49     for(int j=0;j<Q[i].size();j++)50         if(vis[Q[i][j]]&&(Hash[i][Q[i][j]]))    {51             Ans[anc[find(Q[i][j])]]+=Hash[i][Q[i][j]];52             Hash[i][Q[i][j]] = Hash[Q[i][j]][i] = 0;53         }54 }55 56 int main(){57     while(scanf("%d",&n)!=EOF){58         memset(Ans,0,sizeof(Ans));Es = 0;59         memset(Hash,false,sizeof(Hash));60         memset(head,-1,sizeof(head));61         memset(indeg,0,sizeof(indeg));memset(anc,0,sizeof(anc));62         memset(vis,false,sizeof(vis));63         For(i,n){64             Q[i].clear();65             x = read(); tot = read();66             For(j,tot){67                 y = read();68                 makelist(x,y);69                 indeg[y]++;70             }71         }72         m = read();73         For(i,m){74             int x = read() , y = read();75             Hash[x][y]++;Hash[y][x]++;76             Q[x].push_back(y);Q[y].push_back(x);77         }78         For(i,n) {79             root = (!indeg[i]) ? i : root;80             fa[i] = i;81         }82         LCA(root);83         For(i,n) 84           if(Ans[i]) printf("%d:%d\n",i,Ans[i]);85     }86     return 0;87 }

 


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