Tarjan模板 求割点数+桥+连通分量数 来自kuangbin

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

/*
*  求 无向图的割点和桥
*  可以找出割点和桥，求删掉每个点后增加的连通块。
*  需要注意重边的处理，可以先用矩阵存，再转邻接表，或者进行判重
*/
const int MAXN = 10010;
const int MAXM = 100010;
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[110][110];
char buf[1010];
int main()
{
int n;
while(scanf("%d",&n)==1 && n)
{
gets(buf);
memset(g,0,sizeof(g));
while(gets(buf))
{
if(strcmp(buf,"0")==0)break;
char *p = strtok(buf," ");
int u;
sscanf(p,"%d",&u);
p = strtok(NULL," ");
int v;
while(p)
{
sscanf(p,"%d",&v);
p = strtok(NULL," ");
g[u][v]=g[v][u]=1;
}
}
init();
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;
}

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

const int MAXN = 110;
const int MAXM = 110*110;

struct Edge
{
int to,next;
}edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
int Index,top;
int scc;
bool Instack[MAXN];

void addedge(int u,int v)
{
edge[tot].to = v;edge[tot].next = head[u];head[u] = tot++;
}
void Tarjan(int u)
{
int v;
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;
for(int i = head[u];i != -1;i = edge[i].next)
{
v = edge[i].to;
if(!DFN[v])
{
Tarjan(v);
if(Low[u] > Low[v])
Low[u] = Low[v];
}
else if(Instack[v] && Low[u] > DFN[v])
Low[u] = DFN[v];
}
if(Low[u] == DFN[u])
{
scc++;
do
{
v = Stack[--top];
Belong[v] = scc;
Instack[v] = false;
}
while( v!= u);
}
}
int in[MAXN],out[MAXN];
void solve(int
dfa9
N)
{
memset(DFN,0,sizeof(DFN));
memset(Instack,false,sizeof(Instack));
Index = scc = top = 0;
for(int i = 1;i <= N;i++)
if(!DFN[i])
Tarjan(i);
if(scc == 1)
{
printf("1\n0\n");
return;
}
for(int i = 1;i <= scc;i++)
in[i] = out[i] = 0;
for(int u = 1;u <= N;u++)
{
for(int i = head[u];i != -1;i = edge[i].next)
{
int v = edge[i].to;
if(Belong[u] != Belong[v])
{
in[Belong[v]]++;
out[Belong[u]]++;
}
}
}
int ans1=0,ans2=0;
for(int i = 1;i <= scc;i++)
{
if(in[i]==0)ans1++;
if(out[i]==0)ans2++;
}
printf("%d\n%d\n",ans1,max(ans1,ans2));

}
void init()
{
tot = 0;
memset(head,-1,sizeof(head));
}
int main()
{
int n;
int v;
while(scanf("%d",&n) == 1)
{
init();
for(int i = 1;i <= n;i++)
{
while(scanf("%d",&v)==1 && v)
{
addedge(i,v);
}
}
solve(n);
}
return 0;
}