poj 3352--Road Construction(双连通分量)

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题意：给定一无向连通图G，问最少再加多少条边使图变成双连通分量（删除任意一条边后任意两点之间仍然可达）。

题解：

易知求出图中各个极大双连通分量，对其进行缩点（所有low值相同的点）。
缩点以后形成的新图G‘，易知G’为连通图，若图中所有点度数均为2，可知图G‘满足双连通分量条件，所以我们统计G’中度数为1的点，假设其个数为n，用归纳法可知，连接这n个点至少需要（n+1）/2（向下取整）条边。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;class node{public:    int x;    node* next;    node():next(0){}};class adjList{private:    int N;    node** adj;public:    adjList(int n):N(n),adj(new node*[n])    {        memset(adj,0,sizeof(node*)*(N));    }    void adjInsert(int a,int b)    {        node*newNode = new node;        newNode->x = b;        newNode->next = adj[a];        adj[a] = newNode;    }    node* adjNode(int x)    {        return adj[x];    }};class solve{private:    int N,R;    adjList* graph;    int* DFN;    int* low;    int* bccIndex;    int* degree;    int index;public:    solve(int n,int r):N(n),R(r),index(1)    {        graph = new adjList(N+5);        DFN = new int[N+5];        low = new int[N+5];        degree = new int[N+5];        memset(DFN,0,sizeof(int)*(N+1));        memset(degree,0,sizeof(int)*(N+5));        processIn();        tarjan(1,0);        calcDegree();        addEdge();    }    void processIn();    void tarjan(int v,int u);    void calcDegree();    void addEdge();};void solve::processIn(){    int a,b;    while(R--)    {        scanf("%d%d",&a,&b);        graph->adjInsert(a,b);        graph->adjInsert(b,a);    }    return ;}void solve::tarjan(int v,int u){    DFN[v] = low[v] = index++;    for(node* tmpNode = graph->adjNode(v);tmpNode != NULL;tmpNode = tmpNode->next)    {        int w = tmpNode->x;        if(w != u&&w != v)        {            if(!DFN[w])     //树边            {                tarjan(w,v);                low[v] = min(low[v],low[w]);            }            else if(DFN[w] < DFN[v])    //后向边            {                low[v] = min(low[v],DFN[w]);            }        }    }    return ;}void solve::calcDegree(){    node* tmpNode;    for(int u = 1;u <= N;u++)    {    for(tmpNode = graph->adjNode(u);tmpNode != NULL;tmpNode = tmpNode->next)    {        int v = tmpNode->x;        if(v != u&&low[u] != low[v])    //low值一样的都在同一个双连通分量        {            degree[low[u]]++;            degree[low[v]]++;        }    }    }    return ;}void solve::addEdge(){    int num = 0;    for(int i = 1;i <= N;i++)    {        if(degree[i] == 2)      //无向图每边计算了两次        {            num++;        }    }    printf("%d\n",(num+1)>>1);  //使叶节点连通的最小边数    return ;}int main(){    int n,r;    while(~scanf("%d%d",&n,&r))    {        solve poj_3352(n,r);    }    return 0;}

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