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:
1 #include<iostream> 2 #include<cstdio> 3 #include<stack> 4 #include<set> 5 6 using namespace std; 7 8 int daan,n,m,head[10001],tot; 9 int dfn[10001],low[10001],sum,rp; 10 int belong[10001],_head[10001],num[10001]; 11 bool vis[10001]; 12 struct kk{ 13 int to,next; 14 }e[50001],a[50001]; 15 stack<int> s; 16 set<int> ans[10001]; 17 18 inline void add(int x,int y) { 19 e[++tot].to = y; 20 e[tot].next = head[x]; 21 head[x] = tot; 22 } 23 24 inline void tarjan(int x) { 25 dfn[x] = low[x] = ++sum; 26 int v; 27 s.push(x); 28 vis[x] = 1; 29 for(int i = head[x];i != 0; i = e[i].next) { 30 v = e[i].to; 31 if(!dfn[v]) { 32 tarjan(v); 33 low[x] = min(low[x],low[v]); 34 } 35 else if(vis[x]) 36 low[x] = min(low[x],dfn[v]); 37 } 38 if(dfn[x] == low[x]) { 39 rp++; 40 do { 41 v = s.top(); 42 s.pop(); 43 belong[v] = rp; 44 num[rp]++; 45 vis[x] = false; 46 } 47 while(v != x); 48 } 49 } 50 51 inline void _add(int x,int y) { 52 a[++tot].to = y; 53 a[tot].next = _head[x]; 54 _head[x] = tot; 55 } 56 57 inline void findin() { 58 for(int i = 1;i <= rp; i++) 59 for(int j = _head[i];j != 0; j = a[j].next) { 60 int o = a[j].to; 61 ans[i].insert(o); 62 } 63 } 64 65 int main() { 66 scanf("%d%d",&n,&m); 67 for(int i = 1;i <= m; i++) { 68 int x,y; 69 scanf("%d%d",&x,&y); 70 add(x,y); 71 } 72 for(int i = 1;i <= n; i++) 73 if(!dfn[i]) //求强连通分量 74 tarjan(i); 75 tot = 0; 76 for(int i = 1;i <= n; i++)//构建新图,其实没啥必要 77 for(int j = head[i];j != 0; j = e[j].next) 78 if(belong[i] != belong[e[j].to]) 79 _add(belong[i],belong[e[j].to]); 80 findin(); 81 for(int i = 1;i <= rp; i++)//找入度为0的点 82 if(ans[i].size() == 0) { 83 if(daan != 0) { 84 printf("0"); 85 return 0; 86 } 87 daan = i; 88 } 89 printf("%d",num[daan]); 90 return 0; 91 }
1 #include<iostream> 2 #include<cstdio> 3 #include<stack> 4 #include<set> 5 6 using namespace std; 7 8 int daan,n,m,head[10001],tot; 9 int _out[10001],dfn[10001],low[10001]; 10 int sum,rp,belong[10001]; 11 int _head[10001],num[10001]; 12 bool vis[10001]; 13 struct kk{ 14 int to,next; 15 }e[50001],a[50001]; 16 stack<int> s; 17 18 inline void add(int x,int y) { 19 e[++tot].to = y; 20 e[tot].next = head[x]; 21 head[x] = tot; 22 } 23 24 inline void tarjan(int x) { 25 dfn[x] = low[x] = ++sum; 26 int v; 27 s.push(x); 28 vis[x] = 1; 29 for(int i = head[x];i != 0; i = e[i].next) { 30 v = e[i].to; 31 if(!dfn[v]) { 32 tarjan(v); 33 low[x] = min(low[x],low[v]); 34 } 35 else if(vis[x]) 36 low[x] = min(low[x],dfn[v]); 37 } 38 if(dfn[x] == low[x]) { 39 rp++; 40 do { 41 v = s.top(); 42 s.pop(); 43 belong[v] = rp; 44 num[rp]++; 45 vis[x] = false; 46 } 47 while(v != x); 48 } 49 } 50 51 inline void _add(int x,int y) { 52 a[++tot].to = y; 53 a[tot].next = _head[x]; 54 _head[x] = tot; 55 } 56 57 int main() { 58 scanf("%d%d",&n,&m); 59 for(int i = 1;i <= m; i++) { 60 int x,y; 61 scanf("%d%d",&x,&y); 62 add(x,y); 63 } 64 for(int i = 1;i <= n; i++) 65 if(!dfn[i]) 66 tarjan(i); 67 tot = 0; 68 for(int i = 1;i <= n; i++) 69 for(int j = head[i];j != 0; j = e[j].next) 70 if(belong[i] != belong[e[j].to]) 71 _out[belong[i]]++; 72 for(int i = 1;i <= rp; i++) 73 if(_out[i] == 0) { 74 if(daan != 0) { 75 printf("0"); 76 return 0; 77 } 78 daan = i; 79 } 80 printf("%d",num[daan]); 81 return 0; 82 }