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  • POJ3694 Network(连通图+LCA)

    题目链接:http://poj.org/problem?id=3694

    题目大意:给定一个图,每次添加一条边(可能有重边)。输出每次添加后桥的

    数目。由于添加边的次数比较多,添加一次Tarjin一次明显会超时。后来查到了

    LCA算法,利用保存的子节点与最近父节点的关系进行计算的。第一次Tarjin后将

    桥和其所在的父子节点关系保存下来,之后的m次添加边只需要在LCA中判断添加

    的边是否为桥,若为桥则将此桥标记抹去,并将桥数减一,否则无影响。

    附AC代码:

    #include <stdio.h>
    #include <algorithm>
    #include <string.h>
    using namespace std;
    struct ad
    {
        int to, next, used;
    }edge[400005];
    int head[100005], edge_num, dfn[100005], low[100005], bridge_num, Time, father[100005];
    bool isbridge[100005];
    void Add(int x, int y)
    {
        edge[edge_num].to = y;
        edge[edge_num].next = head[x];
        edge[edge_num].used = 0;
        head[x] = edge_num++;
    }
    void Init()
    {
        memset(head, -1, sizeof(head));
        memset(father, 0, sizeof(father));
        memset(low, 0, sizeof(low));
        memset(dfn, 0, sizeof(dfn));
        memset(isbridge, false, sizeof(isbridge));
        edge_num = Time = bridge_num = 0;
    }
    void Tarjin(int u, int fa)
    {
        father[u] = fa;
        low[u] = dfn[u] = ++Time;
        int i, v;
        for(i=head[u]; i!=-1; i=edge[i].next)
        {
            if(edge[i].used==0)
            {
                edge[i].used = edge[i^1].used = 1;
                v = edge[i].to;
                if(!dfn[v])
                {
                    Tarjin(v, u);
                    low[u] = min(low[u], low[v]);
                    if(low[v]>dfn[u])
                    {
                        bridge_num++;
                        isbridge[v] = true;
                    }
                }
                else if(v!=fa)
                    low[u] = min(low[u], dfn[v]);
            }
        }
    }
    void LCA(int u, int v)
    {
        if(dfn[u]>dfn[v])
            swap(u, v);
        while(dfn[v]>dfn[u])
        {
            if(isbridge[v])bridge_num--;
            isbridge[v] = false;
            v = father[v];
        }
    
        while(u!=v)
        {
            if(isbridge[u])bridge_num--;
            if(isbridge[v])bridge_num--;
            isbridge[u] = isbridge[v] = false;
            u = father[u];
            v = father[v];
        }
    }
    int main()
    {
        int n, m, icase = 1;
        while(scanf("%d %d", &n, &m), n+m)
        {
            int a, b;
            Init();
            for(int i=1; i<=m; i++)
            {
                scanf("%d %d", &a, &b);
                Add(a, b);
                Add(b, a);
            }
            Tarjin(1, -1);
            int Q;
            scanf("%d", &Q);
            printf("Case %d:
    ", icase++);
            while(Q--)
            {
                scanf("%d %d", &a, &b);
                LCA(a, b);
                printf("%d
    ", bridge_num);
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zznulw/p/5785260.html
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