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  • hdu 4612 边连通度缩点+树的最长路径

    思路:将以桥为分界的所有连通分支进行缩点,得到一颗树,求出树的直径。再用树上的点减去直径,再减一

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<map>
    #include<queue>
    #define Maxn 210110
    #define Maxm 2501000
    using namespace std;
    int index[Maxn],vi[Maxn],dfn[Maxn],low[Maxn],e,n,lab=0,Stack[Maxn],top,num,head[Maxn],ans,id[Maxn];
    void init()
    {
        memset(index,-1,sizeof(index));
        memset(head,-1,sizeof(head));
        memset(vi,0,sizeof(vi));
        memset(low,0,sizeof(low));
        memset(dfn,0,sizeof(dfn));
        e=lab=top=num=ans=0;
    }
    struct Edge{
        int from,to,next,v;
    }edge[Maxm];
    void addedge(int from, int to)
    {
        edge[e].v=0;
        edge[e].from=from;
        edge[e].to=to;
        edge[e].next=index[from];
        index[from]=e++;
        edge[e].v=0;
        edge[e].to=from;
        edge[e].from=to;
        edge[e].next=index[to];
        index[to]=e++;
    }
    void add(int from,int to)
    {
        edge[e].v=0;
        edge[e].from=from;
        edge[e].to=to;
        edge[e].next=head[from];
        head[from]=e++;
        edge[e].v=0;
        edge[e].to=from;
        edge[e].from=to;
        edge[e].next=head[to];
        head[to]=e++;
    }
    int Count(int u)
    {
        ++num;
        int i;
        do
        {//将该连通分量进行标记
            i=Stack[--top];
            id[i]=num;
        }while(i!=u);
        return 0;
    }
    int dfs(int u)
    {
        dfn[u]=low[u]=++lab;
        Stack[top++]=u;
        int i,j,temp;
        for(i=index[u];i!=-1;i=edge[i].next)
        {
            temp=edge[i].to;
            if(edge[i].v) continue;//一开始没加这个判断,一直WA
            edge[i].v=edge[i^1].v=1;
            if(!dfn[temp])
            {
                dfs(temp);
                low[u]=min(low[u],low[temp]);
            }
            low[u]=min(low[u],dfn[temp]);
        }
        if(dfn[u]==low[u])
            Count(u);
        return 0;
    }
    int maxLen(int u)
    {
        vi[u]=1;
        int i,j,temp=0,Max=0,lMax=0;
        for(i=head[u];i!=-1;i=edge[i].next)
        {
            if(!vi[edge[i].to])
            {
                temp=maxLen(edge[i].to);
                if(temp+1>=Max)
                {
                    lMax=Max;
                    Max=temp+1;
                }
                else
                {
                    if(temp+1>lMax)
                        lMax=temp+1;
                }
                if(Max+lMax>ans)
                    ans=Max+lMax;
            }
        }
        return Max;
    }
    int solve()
    {
        int i,j,u,v,ed;
        dfs(1);
        ed=e;
        e=0;
        for(i=0;i<ed;i+=2)
        {
            u=id[edge[i].from];
            v=id[edge[i].to];
            add(u,v);
        }
        memset(vi,0,sizeof(vi));
        ans=0;
        maxLen(1);
        return num-1-ans;
    }
    int main()
    {
        int m,i,j,a,b;
        while(scanf("%d%d",&n,&m),n||m)
        {
            init();
            for(i=1;i<=m;i++)
            {
                scanf("%d%d",&a,&b);
                addedge(a,b);
            }
            printf("%d
    ",solve());
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/wangfang20/p/3215313.html
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