BNUOJ 1589 Closest Common Ancestors

Closest Common Ancestors

Time Limit: 2000ms

Memory Limit: 10000KB

This problem will be judged on PKU. Original ID: 1470
64-bit integer IO format: %lld      Java class name: Main

 

 

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

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

Huge input, scanf is recommended.

 

Source

Southeastern Europe 2000

 

解题：LCA。。。。

 

#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 = 1000;
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 = 0; 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 = 0; 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 = 0; i <= n; i++) {
            g[i].clear();
            q[i].clear();
            cnt[i] = 0;
            indeg[i] = false;
        }
        for(i = 0; i < n; i++) {
            scanf("%d:(%d)",&u,&k);
            for(j = 0; 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,0,sizeof(cnt));
        for(i = 1; i <= n; i++)
            if(!indeg[i]) {
                tarjan(i);
                break;
            }
        for(i = 1; i <= n; i++)
            if(cnt[i]) printf("%d:%d
",i,cnt[i]);
    }
    return 0;
}