POJ 1470 Closest Common Ancestors (LCA,离线Tarjan算法)
Time Limit: 2000MS | Memory Limit: 10000K | |
Total Submissions: 13372 | Accepted: 4340 |
Description
Input
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 example, for the following tree:
Sample Input
5
5:(3) 1 4 2
1:(0)
4:(0)
2:(1) 3
3:(0)
6
(1 5) (1 4) (4 2)
(2 3)
(1 3) (4 3)
Sample Output
2:1
5:5
Hint
/* ***********************************************
Author :kuangbin
Created Time :2013-9-5 9:11:48
File Name :F:\2013ACM练习\专题学习\LCA\POJ1470_2.cpp
************************************************ */ #include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
/*
* POJ 1470
* 给出一颗有向树,Q个查询
* 输出查询结果中每个点出现次数
*/
/*
* LCA离线算法,Tarjan
* 复杂度O(n+Q);
*/
const int MAXN = ;
const int MAXQ = ;//查询数的最大值 //并查集部分
int F[MAXN];//需要初始化为-1
int find(int x)
{
if(F[x] == -)return x;
return F[x] = find(F[x]);
}
void bing(int u,int v)
{
int t1 = find(u);
int t2 = find(v);
if(t1 != t2)
F[t1] = t2;
}
//************************
bool vis[MAXN];//访问标记
int ancestor[MAXN];//祖先
struct Edge
{
int to,next;
}edge[MAXN*];
int head[MAXN],tot;
void addedge(int u,int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
} struct Query
{
int q,next;
int index;//查询编号
}query[MAXQ*];
int answer[MAXQ];//存储最后的查询结果,下标0~Q-1
int h[MAXQ];
int tt;
int Q; void add_query(int u,int v,int index)
{
query[tt].q = v;
query[tt].next = h[u];
query[tt].index = index;
h[u] = tt++;
query[tt].q = u;
query[tt].next = h[v];
query[tt].index = index;
h[v] = tt++;
} void init()
{
tot = ;
memset(head,-,sizeof(head));
tt = ;
memset(h,-,sizeof(h));
memset(vis,false,sizeof(vis));
memset(F,-,sizeof(F));
memset(ancestor,,sizeof(ancestor));
} void LCA(int u)
{
ancestor[u] = u;
vis[u] = true;
for(int i = head[u];i != -;i = edge[i].next)
{
int v = edge[i].to;
if(vis[v])continue;
LCA(v);
bing(u,v);
ancestor[find(u)] = u;
}
for(int i = h[u];i != -;i = query[i].next)
{
int v = query[i].q;
if(vis[v])
{
answer[query[i].index] = ancestor[find(v)];
}
}
} bool flag[MAXN];
int Count_num[MAXN];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
int u,v,k;
while(scanf("%d",&n) == )
{
init();
memset(flag,false,sizeof(flag));
for(int i = ;i <= n;i++)
{
scanf("%d:(%d)",&u,&k);
while(k--)
{
scanf("%d",&v);
flag[v] = true;
addedge(u,v);
addedge(v,u);
}
}
scanf("%d",&Q);
for(int i = ;i < Q;i++)
{
char ch;
cin>>ch;
scanf("%d %d)",&u,&v);
add_query(u,v,i);
}
int root;
for(int i = ;i <= n;i++)
if(!flag[i])
{
root = i;
break;
}
LCA(root);
memset(Count_num,,sizeof(Count_num));
for(int i = ;i < Q;i++)
Count_num[answer[i]]++;
for(int i = ;i <= n;i++)
if(Count_num[i] > )
printf("%d:%d\n",i,Count_num[i]);
}
return ;
}
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