记录一个菜逼的成长。。
题目链接
题目大意:
判断给你的图是否为强连通图
tarjan算法套下就好了
判断强连通分量的数量是否为1
#include <bits/stdc++.h>
using namespace std;
#define cl(a,b) memset(a,b,sizeof(a))
const int maxn = 100000 + 10;
int head[maxn],cnt;
struct Edge{
int next,to;
}edge[maxn];
void add(int u,int v)
{
edge[cnt].to = v;
edge[cnt].next = head[u];
head[u] = cnt++;
}
int instack[maxn],dfn[maxn],low[maxn],ind,num,belong[maxn];
stack<int>sta;
void tarjan(int u)
{
///dfn(u)为时间戳,即dfs序
///Low(u)为u或u的子树能够追溯到的最早的栈中节点的次序号
dfn[u] = low[u] = ++ind;
instack[u] = 1;
sta.push(u);
for( int i = head[u]; ~i; i = edge[i].next ){
int v = edge[i].to;
if(!dfn[v]){
tarjan(v);
low[u] = min(low[u],low[v]);
}
else if(instack[v]){
low[u] = min(low[u],dfn[v]);
}
}
if(dfn[u] == low[u]){
num++;
int v;
do{
v = sta.top();
sta.pop();
instack[v] = 0;
belong[v] = num;///点v属于第num个强连通分量
}while(u != v);
}
}
void init()
{
while(!sta.empty())sta.pop();
cl(head,-1),cnt = 0;
cl(instack,0),cl(dfn,0),cl(low,0);
ind = 0;
num = 0;
cl(belong,0);
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m),n||m){
init();
for( int i = 0; i < m; i++ ){
int u,v;
scanf("%d%d",&u,&v);
add(u,v);
}
for( int i = 1; i <= n; i++ )
if(!dfn[i])tarjan(i);
puts(num!=1?"No":"Yes");
}
return 0;
}