HDU 2767 Proving Equivalences 强连通分量

题目：http://acm.hdu.edu.cn/showproblem.php?pid=2767

题意：题目描述很繁杂，大概意思就是求最少加几条边可以使图只有一个强连通分量。

思路：用tarjan算法求强连通分量缩点，统计每个点的入度和出度，最后输出入度为0和出度为0的点的个数中的较大值，至于为什么这样做，没有证明，不过可以自己实验一下

总结：跟poj某道题差不多

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
#include<cstring>
#include<cctype>
#include<cmath>
using namespace std;
typedef long long ll;
const int N = 20100;
struct edge
{
int to, next;
} G[N*5];
int dfn
, low
, scc
, st
, head
;
int index, cnt, top, num;
bool vis
;
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++;
vis[v] = true;
st[top++] = v;
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 outdeg
, indeg
;
memset(outdeg, 0, sizeof outdeg);
memset(indeg, 0, sizeof indeg);
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(outdeg[i] == 0) out0++;
if(indeg[i] == 0) in0++;
}
printf("%d\n", max(in0, out0));
}
int main()
{
int t, a, b;
scanf("%d", &t);
while(t--)
{
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;
}