洛谷 P2341 [HAOI2006]受欢迎的牛|【模板】强连通分量

题目传送门

解题思路:

先求强联通分量,缩点,然后统计新图中有几个点出度为0,如果大于1个,则说明这不是一个连通图,答案即为0.否则入度为0的那个强连通分量的点数即为答案

AC代码:

 #include<iostream>
 #include<cstdio>
 #include<stack>
 #include<set> 

 using namespace std;

 int daan,n,m,head[],tot;
 int dfn[],low[],sum,rp;
 int belong[],_head[],num[];
 bool vis[];
 struct kk{
     int to,next;
 }e[],a[];
 stack<int> s;
 set<int> ans[];

 inline void add(int x,int y) {
     e[++tot].to = y;
     e[tot].next = head[x];
     head[x] = tot;
 }

 inline void tarjan(int x) {
     dfn[x] = low[x] = ++sum;
     int v;
     s.push(x);
     vis[x] = ;
     for(int i = head[x];i != ; i = e[i].next) {
         v = e[i].to;
         if(!dfn[v]) {
             tarjan(v);
             low[x] = min(low[x],low[v]);
         }
         else if(vis[x])
             low[x] = min(low[x],dfn[v]);
     }
     if(dfn[x] == low[x]) {
         rp++;
         do {
             v = s.top();
             s.pop();
             belong[v] = rp;
             num[rp]++;
             vis[x] = false;
         }
         while(v != x);
     }
 }

 inline void _add(int x,int y) {
     a[++tot].to = y;
     a[tot].next = _head[x];
     _head[x] = tot;
 }

 inline void findin() {
     for(int i = ;i <= rp; i++)
         for(int j = _head[i];j != ; j = a[j].next) {
             int o = a[j].to;
             ans[i].insert(o);
         }
 } 

 int main() {
     scanf("%d%d",&n,&m);
     for(int i = ;i <= m; i++) {
         int x,y;
         scanf("%d%d",&x,&y);
         add(x,y);
     }
     for(int i = ;i <= n; i++)
         if(!dfn[i])    //求强连通分量
             tarjan(i);
     tot = ;
     for(int i = ;i <= n; i++)//构建新图,其实没啥必要
         for(int j = head[i];j != ; j = e[j].next)
             if(belong[i] != belong[e[j].to])
                 _add(belong[i],belong[e[j].to]);
     findin();
     for(int i = ;i <= rp; i++)//找入度为0的点
         if(ans[i].size() == ) {
             if(daan != )  {
                 printf("");
                 return ;
             }
             daan = i;
         }
     printf("%d",num[daan]);
     return ;
 } 

第一次的做法,用了set,较麻烦

 #include<iostream>
 #include<cstdio>
 #include<stack>
 #include<set> 

 using namespace std;

 int daan,n,m,head[],tot;
 int _out[],dfn[],low[];
 int sum,rp,belong[];
 int _head[],num[];
 bool vis[];
 struct kk{
     int to,next;
 }e[],a[];
 stack<int> s;

 inline void add(int x,int y) {
     e[++tot].to = y;
     e[tot].next = head[x];
     head[x] = tot;
 }

 inline void tarjan(int x) {
     dfn[x] = low[x] = ++sum;
     int v;
     s.push(x);
     vis[x] = ;
     for(int i = head[x];i != ; i = e[i].next) {
         v = e[i].to;
         if(!dfn[v]) {
             tarjan(v);
             low[x] = min(low[x],low[v]);
         }
         else if(vis[x])
             low[x] = min(low[x],dfn[v]);
     }
     if(dfn[x] == low[x]) {
         rp++;
         do {
             v = s.top();
             s.pop();
             belong[v] = rp;
             num[rp]++;
             vis[x] = false;
         }
         while(v != x);
     }
 }

 inline void _add(int x,int y) {
     a[++tot].to = y;
     a[tot].next = _head[x];
     _head[x] = tot;
 }

 int main() {
     scanf("%d%d",&n,&m);
     for(int i = ;i <= m; i++) {
         int x,y;
         scanf("%d%d",&x,&y);
         add(x,y);
     }
     for(int i = ;i <= n; i++)
         if(!dfn[i])
             tarjan(i);
     tot = ;
     for(int i = ;i <= n; i++)
         for(int j = head[i];j != ; j = e[j].next)
             if(belong[i] != belong[e[j].to])
                 _out[belong[i]]++;
     for(int i = ;i <= rp; i++)
         if(_out[i] == ) {
             if(daan != )  {
                 printf("");
                 return ;
             }
             daan = i;
         }
     printf("%d",num[daan]);
     return ;
 } 

改进后的代码,发现不用set,比较精简