POJ-1470-Closest Common Ancestors 解题报告

来源：互联网发布：linux vps 教程编辑：程序博客网时间：2024/05/16 01:13

LCA简单基础入门题。赤裸裸的求LCA的题，不过输入比较麻烦。题意：其实看图加样例就能懂了，不多说了，最后输出每个节点成为括号内节点LCA的节点id和次数，次数为0就不输出了。

我的解题思路：用Tarjan的离线算法最后把答案哈希输出就ok了，关于此算法的原理等见请看上一个解题报告。

我的解题代码：离线求LCA算法

#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>#include <vector>using namespace std;#define N 1000bool root[N];       //判断是否根节点bool vis[N];int father[N];      //幷查集使用int lca[N];         //LCAint ans[N];vector <int> e[N];  //存储子节点vector <int> q[N];  //存储询问int n, qn;void InitRead();int Find(int x);void DataProcess(int x);int main(){    while (~scanf("%d", &n))    {        InitRead();        for (int i=1; i<=n; ++i)        {            if (root[i])            {                DataProcess(i);                break;            }        }        for (int i=1; i<=n; ++i)        {            if (ans[i] != 0)            {                printf("%d:%d\n", i, ans[i]);            }        }    }    return 0;}void InitRead(){    for (int i=0; i<=n; ++i)    {        root[i] = true;        vis[i] = false;        father[i] = i;        ans[i] = lca[i] = 0;        e[i].clear();        q[i].clear();    }    int a, b, c;    for (int i=0; i<n; ++i)    {        scanf("%d:(%d)", &a, &b);        while (b--)        {            scanf("%d", &c);            e[a].push_back(c);            root[c] = false;        }    }    scanf("%d", &qn);    while (qn--)    {        scanf(" (%d %d)", &a, &b);        q[a].push_back(b);        q[b].push_back(a);    }    return;}int Find(int x){    int z, y = x;    while (y != father[y]) y = father[y];    while (x != father[x])    {        z = father[x];        father[x] = y;        x = z;    }    return y;}void DataProcess(int x){    lca[x] = x;    int size = e[x].size();    for (int i=0; i<size; ++i)    {        DataProcess(e[x][i]);        father[e[x][i]] = x;    }    vis[x] = true;    size = q[x].size();    for (int i=0; i<size; ++i)    {        if (vis[q[x][i]]) ans[lca[Find(q[x][i])]]++;    //把答案哈希    }    return;}

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