BNUOJ 1589 Closest Common Ancestors
Closest Common Ancestors
This problem will be judged on PKU. Original ID: 1470
64-bit integer IO format: %lld Java class name: Main
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
Source
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <climits>
#include <algorithm>
#include <cmath>
#define LL long long
#define INF 0x3f3f3f
using namespace std;
const int maxn = ;
vector<int>g[maxn];
vector<int>q[maxn];
int n,m,cnt[maxn],uf[maxn];
bool vis[maxn],indeg[maxn];
int Find(int x) {
if(x != uf[x])
uf[x] = Find(uf[x]);
return uf[x];
}
void tarjan(int u) {
int i;
uf[u] = u;
for(i = ; i < g[u].size(); i++) {
if(!vis[g[u][i]] && g[u][i] != u) {
tarjan(g[u][i]);
uf[g[u][i]] = u;
}
}
vis[u] = true;
for(i = ; i < q[u].size(); i++) {
if(vis[q[u][i]]) cnt[Find(q[u][i])]++;
}
}
int main() {
int i,j,u,v,k;
while(~scanf("%d",&n)) {
for(i = ; i <= n; i++) {
g[i].clear();
q[i].clear();
cnt[i] = ;
indeg[i] = false;
}
for(i = ; i < n; i++) {
scanf("%d:(%d)",&u,&k);
for(j = ; j < k; j++) {
scanf("%d",&v);
g[u].push_back(v);
indeg[v] = true;
}
}
scanf("%d",&m);
while(m--) {
scanf(" (%d %d)",&u,&v);
q[u].push_back(v);
q[v].push_back(u);
}
memset(vis,false,sizeof(vis));
memset(cnt,,sizeof(cnt));
for(i = ; i <= n; i++)
if(!indeg[i]) {
tarjan(i);
break;
}
for(i = ; i <= n; i++)
if(cnt[i]) printf("%d:%d\n",i,cnt[i]);
}
return ;
}
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