• uva315(求割点数目)


    传送门:Network

    题意:给出一张无向图,求割点的个数。

    分析:模板裸题,直接上模板。

    #include <cstdio>
    #include <cstring>
    #include <string>
    #include <cmath>
    #include <iostream>
    #include <algorithm>
    #include <queue>
    #include <cstdlib>
    #include <stack>
    #include <vector>
    #include <set>
    #include <map>
    #define LL long long
    #define mod 100000000
    #define inf 0x3f3f3f3f
    #define eps 1e-6
    #define N 110
    #define FILL(a,b) (memset(a,b,sizeof(a)))
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define PII pair<int,int>
    using namespace std;
    struct edge
    {
        int v,next;
        edge(){}
        edge(int v,int next):v(v),next(next){}
    }e[N*N];
    int n,step,top,tot;
    int head[N],dfn[N],low[N],Stack[N];
    bool instack[N],cutpoint[N];
    void init()
    {
        tot=0;step=0;top=0;
        FILL(head,-1);FILL(dfn,0);
        FILL(low,0);FILL(instack,false);
        FILL(cutpoint,false);
    }
    void addedge(int u,int v)
    {
        e[tot]=edge(v,head[u]);
        head[u]=tot++;
    }
    void tarjan(int u,int fa)
    {
        int son=0;
        dfn[u]=low[u]=++step;
        Stack[top++]=u;
        instack[u]=true;
        for(int i=head[u];~i;i=e[i].next)
        {
            int v=e[i].v;
            if(v==fa)continue;
            if(!dfn[v])
            {
                son++;
                tarjan(v,u);
                if(low[u]>low[v])low[u]=low[v];
                //割点: 一个顶点u是割点,当且仅当满足(1)或(2)
                //(1) u为树根,且u有多于一个子树
                //(2) u不为树根,且满足存在(u,v)为树枝边,使得DFS(u)<=Low(v)
                if(u!=fa&&low[v]>=dfn[u])
                {
                    cutpoint[u]=true;
                }
            }
            else if(low[u]>dfn[v])
            {
                low[u]=dfn[v];
            }
        }
        //树根,分支数大于1
        if(u==fa&&son>1)cutpoint[u]=true;
        instack[u]=false;
        top--;
    }
    void solve()
    {
        for(int i=1;i<=n;i++)
            if(!dfn[i])tarjan(i,i);
        int ans=0;
        for(int i=1;i<=n;i++)
            if(cutpoint[i])ans++;
        printf("%d
    ",ans);
    }
    int main()
    {
        int u,v;
        while(scanf("%d",&n),n)
        {
            init();
            while(scanf("%d",&u),u)
            {
                while(getchar()!='
    ')
                {
                    scanf("%d",&v);
                    addedge(u,v);
                    addedge(v,u);
                }
            }
            solve();
        }
    }
    View Code
  • 相关阅读:
    poj2386 Lake Counting
    poj 1852 Ants
    Luogu P2419 [USACO08JAN]牛大赛Cow Contest
    Luogu P2336 [SCOI2012]喵星球上的点名
    Luogu P2463 [SDOI2008]Sandy的卡片
    Luogu P2852 [USACO06DEC]牛奶模式Milk Patterns
    Luogu P4248 [AHOI2013]差异
    【NOI2008】志愿者招募
    Luogu P2743 [USACO5.1]乐曲主题Musical Themes
    P3723 [AH2017/HNOI2017]礼物
  • 原文地址:https://www.cnblogs.com/lienus/p/4278786.html
Copyright © 2020-2023  润新知