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  • 最小公共祖先 (Tarjan) POJ1470

    POJ 1470

    标准的LCA,输入感觉怪怪的=.=

    自己看了下Tarjan,再参考了下别人的处理方法(感觉自己好弱。。)


    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    using namespace std;
    
    const int N = 1010;
    const int M = 1000010;
    
    int in[N], par[N], ans[N];
    bool vis[N];
    int n, edge, m;
    
    struct ST
    {
        int top;
        int next[M],key[M],head[N];
        void add(int x,int y)
        {
            key[top] = y;       //记录儿子
            next[top] = head[x];          
            head[x] = top++;
        }
        void clear()
        {
            top = 0;
            memset(head,-1,sizeof(head));
        }
    }son,q;
    
    int fin(int x)
    {
        if(x != par[x])
            par[x] = fin(par[x]);
        return par[x];
    }
    
    void lca(int cur)
    {
        par[cur] = cur;
        for(int i = son.head[cur];i != -1;i = son.next[i])
        {
            lca(son.key[i]);
            par[son.key[i]] = cur;
        }
        vis[cur] = true;
        for(int i = q.head[cur];i!= -1;i = q.next[i])
        {
            if(vis[q.key[i]])
                ans[fin(q.key[i])]++;
        }
    }
    
    void ini()
    {
        memset(in,0,sizeof(in));
        memset(vis,false,sizeof(vis));
        memset(ans,0,sizeof(ans));
    }
    
    int main()
    {
        int root, u, num, v;
        while(scanf("%d", &n) != EOF)
        {
            ini();
            son.clear(), q.clear();
            for(int j = 0; j < n; ++j)
            {
                scanf("%d:(%d)", &u, &num);
                for(int i = 0; i < num; ++i)
                {
                    scanf("%d", &v);
                    son.add(u, v);
                    in[v]++;
                }
            }
        for(root = 1;in[root];root++);
        scanf("%d",&m);
        while(m--)
            {
                scanf(" (%d %d)", &u, &v);
                if(u == v)
                    q.add(v, u);
                else
                {
                    q.add(u, v);
                    q.add(v, u);
                }
            }
            lca(root);
            for(int i = 1;i <= n;i++)
                if(ans[i])
                 printf("%d:%d
    ", i, ans[i]);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Przz/p/5409841.html
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