POJ 1470 Closest Common Ancestors 【LCA】
任意门:http://poj.org/problem?id=1470
Time Limit: 2000MS | Memory Limit: 10000K | |
Total Submissions: 22519 | Accepted: 7137 |
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
题意概括:
给一棵 N 个结点 N-1 条边的树,和 M 次查询;
形式是:先给根结点 然后 给与这个根结点相连的 子节点。
查询的对儿给得有点放荡不羁,需要处理一下空格、回车、括号。。。
解题思路:
虽说表面上看是一道裸得 LCA 模板题(简单粗暴Tarjan)
但是细节还是要注意:
本题没有给 查询数 M 的范围(RE了两次)所以要投机取巧一下不使用记录每对查询的 LCA。
本题是多测试样例,注意初始化!!!
AC code:
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int MAXN = 1e3+;
struct Edge{int v, w, nxt;}edge[MAXN<<];
struct Query
{
int v, id;
Query(){};
Query(int _v, int _id):v(_v), id(_id){};
};
vector<Query> q[MAXN]; int head[MAXN], cnt;
int fa[MAXN], ans[MAXN<<], no[MAXN];
bool vis[MAXN], in[MAXN];
int N, M; void init()
{
memset(head, -, sizeof(head));
memset(in, false, sizeof(in));
memset(vis, false, sizeof(vis));
//memset(ans, 0, sizeof(ans));
memset(no, , sizeof(no));
cnt = ;
for(int i = ; i <= N; i++)
fa[i] = i, q[i].clear();
} int getfa(int x){return fa[x]==x?x:fa[x]=getfa(fa[x]);} void AddEdge(int from, int to)
{
edge[cnt].v = to;
edge[cnt].nxt = head[from];
head[from] = cnt++;
} void Tarjan(int s, int f)
{
int root = s;
fa[s] = s;
for(int i = head[s]; i != -; i = edge[i].nxt)
{
int Eiv = edge[i].v;
if(Eiv == f) continue;
Tarjan(Eiv, s);
fa[getfa(Eiv)] = s;
}
vis[s] = true;
for(int i = ; i < q[s].size(); i++){
//if(vis[q[s][i].v]) ans[q[s][i].id] = getfa(q[s][i].v);
if(vis[q[s][i].v])
no[getfa(q[s][i].v)]++;
}
} int main()
{
while(~scanf("%d", &N)){
init();
//scanf("%d", &N);
for(int i = , u, v, k; i <= N; i++){
scanf("%d:(%d)", &u, &k);
for(int j = ; j <= k; j++){
scanf("%d", &v);
AddEdge(u, v);
in[v] = true;
}
}
int root = ;
for(int i = ; i <= N; i++) if(!in[i]){root = i;break;}
scanf("%d", &M);
int sum = , u, v;
while(sum <= M){
while(getchar()!='(');
scanf("%d%d", &u, &v);
while(getchar()!=')');
q[u].push_back(Query(v, sum));
q[v].push_back(Query(u, sum));
sum++;
}
Tarjan(root, -);
/*
for(int i = 1; i <= M; i++){
no[ans[i]]++;
}
*/ for(int i = ; i <= N; i++){
if(no[i]) printf("%d:%d\n", i, no[i]);
}
}
return ;
}
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