ccf模拟题 201709-4 通信网络（缩点，dfs）

有小伙伴问这个问题，就写了一发给他，结果我写的得了95，他仿照我写的得了100，很蛋疼，调了一下，发现tarjan里面写错个变量。


思路：缩点，重新建两个图，正向一个，反向一个，每个点dfs一发即可。O(n^2)

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

const int MAXN = 1010;
const int MAXM = 10100;
struct Edge
{
int to,next;
} edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
int degree[MAXN];
int Index,top;
int scc;
bool Instack[MAXN];
int num[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];
Instack[v] = false;
Belong[v] = scc;
num[scc]++;
}
while(v != u);
}
}

void init()
{
memset(degree,0,sizeof(degree));
memset(DFN,0,sizeof(DFN));
memset(Instack,false,sizeof(Instack));
memset(num,0,sizeof(num));
Index = scc = top = 0;
}

void solve(int n)
{
init();
for(int i = 1; i <= n; ++i)
if(!DFN[i])
Tarjan(i);
}

vector<int> G1[MAXN],G2[MAXN];
bool vis1[MAXN],vis2[MAXN];
int cnt;

void dfs1(int u)
{
cnt++;
for(int i = 0; i < G1[u].size(); ++i)
{
int v = G1[u][i];
if(vis1[v]) continue;
vis1[v] = true;
dfs1(v);
}
}

void dfs2(int u)
{
cnt++;
for(int i = 0; i < G2[u].size(); ++i)
{
int v = G2[u][i];
if(vis2[v]) continue;
vis2[v] = true;
dfs2(v);
}
}
int res;
void rebuild(int n)
{
int u,v;
for(int i = 1; i <= n; ++i)
{
for(int j = head[i]; j != -1; j = edge[j].next)
{
u = Belong[i];
v = Belong[edge[j].to];
if(u != v)
{
G1[u].push_back(v);
G2[v].push_back(u);
}
}
}
res = 0;

for(int i = 1; i <= scc; ++i)
{
memset(vis1,0,sizeof(vis1));
memset(vis2,0,sizeof(vis2));
cnt = 0;
vis1[i] = vis2[i] = true;
dfs1(i);
dfs2(i);
if(cnt == scc+1)
res += num[i];
}
}

int n,m;
int main()
{
scanf("%d %d",&n,&m);
int u,v;
tot = 0;
memset(head,-1,sizeof(head));
for(int i = 0; i < m; ++i)
{
scanf("%d %d",&u,&v);
addedge(u,v);
}
solve(n);
rebuild(n);
printf("%d\n",res);
return 0;
}