ZOj 1141 Closest Common Ancestors（LCA)

最新推荐文章于 2019-08-07 16:17:00 发布

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本文深入探讨了最近公共祖先问题，并提供了一个高效算法的实现。文章详细介绍了求解过程，包括输入数据格式、核心算法逻辑以及关键点注意事项。通过实例分析，帮助读者理解并解决实际问题。

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Closest Common Ancestors

Time Limit:10000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 1141

Appoint description:

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)

The data set 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. The tree description is followed by a list of pairs of vertices, in the form:

nr_of_pairs
(u v) (x y) ...

The input contents several data sets (at least one).

Note that white-spaces (tabs, spaces and line breaks) can be used freely in the input.

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:

the program input and output is:

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)

Output

2:1
5:5

主要就是求最近公共祖先问题；

两个挺坑的地方：

第一MAX要开大，我觉着以后在遇到树MAX就比最大的再加个零

第二多个输入case

第三输入有点小窍门

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
const int MAX = 4000;
vector<int> a[MAX];
int f[MAX],r[MAX],num[MAX];
void dfs(int root,int dep)
{
    r[root] = dep;
    for(vector<int>::iterator it = a[root].begin(); it != a[root].end(); it++)
        dfs(*it,dep+1);
}
int main()
{
    int n,x,y,k;
    while(scanf("%d", &n) != EOF)
    {
    for(int i = 1; i <= n; i++)
    {
        a[i].clear();
        f[i] = 0;
        r[i] = 0;
        num[i] = 0;
    }
    for(int i = 1; i <= n; i++)
    {
        scanf("%d:(%d)", &x,&k);
        while(k--)
        {
            scanf(" %d", &y);
            f[y] = x;
            a[x].push_back(y);
        }
    }
    for(k = 1; k <= n; k++)
        if(f[k] == 0)
            break;
    dfs(k,0);
    scanf(" %d", &k);
    getchar();
    while(k--)
    {

        scanf(" (%d,%d)", &x,&y);
        getchar();
        while(x != y) //注意 不能是f[x] != f[y]，以为公共根节点可能就是x或y
        {
            if(r[x] > r[y])
                x = f[x];
            else if(r[x] < r[y])
                y = f[y];
            else if(x != y)
            {
                x = f[x];
                y = f[y];
            }
        }
        num[x]++;
    }
    for(int i = 1; i <= n; i++)
    {
        if(num[i])
            printf("%d:%d\n",i,num[i]);
    }
    }
    return 0;
}