Network——求割点模板题

题目链接

题解：

求割点模板题

注意输入格式转换

需要考虑重边

 

代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include<vector>
#include <map>
using namespace std;

const int maxn = 5010;//点数
const int maxm = 20010;//边数，因为是无向图，所以这个值要*2

struct Edge
{
    int to,next;
    bool cut;//是否是桥标记
} edge[maxm];
int head[maxn],tot;
int low[maxn],dfn[maxn],Stack[maxn],belong[maxn];//belong数组的值是1~scc
int Index,top;
int scc;//边双连通块数/强连通分量的个数
bool Instack[maxn];
int bridge;//桥的数目
bool cut[maxn];
void addedge(int u,int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    edge[tot].cut=false;
    head[u] = tot++;
}
void Tarjan(int u,int pre)
{
    int v;
    low[u] = dfn[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    int son=0;
    int flag=0;
    for(int i = head[u]; i != -1; i = edge[i].next)
    {
        v = edge[i].to;
        if(v == pre && !flag)//处理重边
        {
            flag++;
            continue;
        }
        if( !dfn[v] )
        {
            son++;
            Tarjan(v,u);
            if( low[u] > low[v] )low[u] = low[v];
            if(low[v] > dfn[u])
            {
                bridge++;
                edge[i].cut = true;
                edge[i^1].cut = true;
            }
            if(u == pre && son > 1)cut[u] = true;
            if(u != pre && low[v] >= dfn[u])cut[u] = true;

        }
        else if( Instack[v] && low[u] > dfn[v] )
            low[u] = dfn[v];
    }

    if(low[u] == dfn[u])
    {
        scc++;
        do
        {
            v = Stack[--top];
            Instack[v] = false;
            belong[v] = scc;
        }
        while( v!=u );
    }

}
void init()
{
    tot = 0;
    memset(head,-1,sizeof(head));
}
void solve(int n)
{
    memset(dfn,0,sizeof(dfn));
    memset(Instack,false,sizeof(Instack));
    memset(cut,0,sizeof cut);
    Index = top = scc = 0;
    bridge = 0;
    for(int i = 1;i <= n;i++)
       if(!dfn[i])
          Tarjan(i,i);
    int ans = 0;
    for(int i = 1;i <= n;i++)
       if(cut[i])
          ans++;
    printf("%d\n",ans);

}
int main()
{
    int n;
    int a,b;
    while(~scanf("%d",&n) && n)
    {
        init();
        memset(g,0,sizeof g);
        while(~scanf("%d",&a) && a)
        {
            while(getchar()!='\n')
            {
                scanf("%d",&b);
                addedge(a,b);
                addedge(b,a);
            }
        }
        solve(n);
    }
    return 0;
}
/*
5
5 1 2 3 4
0
6
2 1 3
5 4 6 2
0
0
*/

 

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
const int maxn = 10010;
const int maxm = 100100;
struct Edge
{
    int to,next;
    bool cut;//是否为桥的标记
}edge[maxm];
int head[maxn],tot;
int low[maxn],dfn[maxn],Stack[maxn];
int Index,top;
bool Instack[maxn];
bool cut[maxn];
int add_block[maxn];//删除一个点后增加的连通块
int bridge;

void addedge(int u,int v)
{
    edge[tot].to = v;edge[tot].next = head[u];edge[tot].cut = false;
    head[u] = tot++;
}

void Tarjan(int u,int pre)
{
    int v;
    low[u] = dfn[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    int son = 0;
    for(int i = head[u];i != -1;i = edge[i].next)
    {
        v = edge[i].to;
        if(v == pre)continue;
        if( !dfn[v] )
        {
            son++;
            Tarjan(v,u);
            if(low[u] > low[v])low[u] = low[v];
            //桥
            //一条无向边(u,v)是桥，当且仅当(u,v)为树枝边，且满足DFS(u)<low(v)。
            if(low[v] > dfn[u])
            {
                bridge++;
                edge[i].cut = true;
                edge[i^1].cut = true;
            }
            //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(或称父子边，
            //即u为v在搜索树中的父亲)，使得DFS(u)<=low(v)
            if(u != pre && low[v] >= dfn[u])//不是树根
            {
                cut[u] = true;
                add_block[u]++;
            }
        }
        else if( low[u] > dfn[v])
             low[u] = dfn[v];
    }
    //树根，分支数大于1
    if(u == pre && son > 1)cut[u] = true;
    if(u == pre)add_block[u] = son - 1;
    Instack[u] = false;
    top--;
}

void solve(int n)
{
    memset(dfn,0,sizeof(dfn));
    memset(Instack,false,sizeof(Instack));
    memset(add_block,0,sizeof(add_block));
    memset(cut,false,sizeof(cut));
    Index = top = 0;
    bridge = 0;
    for(int i = 1;i <= n;i++)
       if(!dfn[i])
          Tarjan(i,i);
    int ans = 0;
    for(int i = 1;i <= n;i++)
       if(cut[i])
          ans++;
    printf("%d\n",ans);
}
void init()
{
    tot = 0;
    memset(head,-1,sizeof(head));
}
int g[105][105];
int main()
{
    int n;
    int a,b;
    while(~scanf("%d",&n) && n)
    {
        init();
        memset(g,0,sizeof g);
        while(~scanf("%d",&a) && a)
        {
            while(getchar()!='\n')
            {
                scanf("%d",&b);
                g[a][b]=g[b][a]=1;
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                if(g[i][j])
                {
                    addedge(i,j);
                    addedge(j,i);
                }
            }
        }
        solve(n);
    }
    return 0;
}
/*
5
5 1 2 3 4
0
6
2 1 3
5 4 6 2
0
0
*/