HDU 3836 Equivalent Sets 强连通分量分解

题目：http://acm.hdu.edu.cn/showproblem.php?pid=3836

题意：就是问加多少个边可以使图只有一个强连通分量，，，跟HDU 2767一样，，，

思路：强连通分量分解缩点，分别统计缩点后入度为0和出度为0的点的个数，输出其中的较大值，注意特判图原本只有一个强连通分量时答案为0的情况，不小心哇了一次。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int N = 20010;
struct edge
{
int to, next;
}G[N*3];
int dfn
, low
, scc
, st
;
int head
;
bool vis
;
int index, top, num, cnt;
int n, m;
void init()
{
memset(head, -1, sizeof head);
memset(dfn, -1, sizeof dfn);
memset(vis, 0, sizeof vis);
index = cnt = top = num = 0;
}
void add_edge(int v, int u)
{
G[cnt].to = u;
G[cnt].next = head[v];
head[v] = cnt++;
}
void tarjan(int v)
{
dfn[v] = low[v] = index++;
st[top++] = v;
vis[v] = true;
int u;
for(int i = head[v]; i != -1; i = G[i].next)
{
u = G[i].to;
if(dfn[u] == -1)
{
tarjan(u);
low[v] = min(low[v], low[u]);
}
else if(vis[u])
low[v] = min(low[v], dfn[u]);
}
if(dfn[v] == low[v])
{
num++;
do
{
u = st[--top];
vis[u] = false;
scc[u] = num;
}while(u != v);
}
}
void slove()
{
for(int i = 1; i <= n; i++)
if(dfn[i] == -1)
tarjan(i);
if(num == 1)
{
printf("0\n");
return;
}
int indeg
, outdeg
;
memset(indeg, 0, sizeof indeg);
memset(outdeg, 0, sizeof outdeg);
for(int i = 1; i <= n; i++)
for(int j = head[i]; j != -1; j = G[j].next)
if(scc[i] != scc[G[j].to])
outdeg[scc[i]]++, indeg[scc[G[j].to]]++;
int in0 = 0, out0 = 0;
for(int i = 1; i <= num; i++)
{
if(indeg[i] == 0) in0++;
if(outdeg[i] == 0) out0++;
}
printf("%d\n", max(in0, out0));
}

int main()
{
int a, b;
while(~ scanf("%d%d", &n, &m))
{
init();
for(int i = 0; i < m; i++)
{
scanf("%d%d", &a, &b);
add_edge(a, b);
}
slove();
}

return 0;
}