传送门：http://acm.hdu.edu.cn/showproblem.php?pid=2767
Proving Equivalences
Time Limit: 4000/2000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
Problem Description
Consider the following exercise, found in a generic linear algebra textbook.
Let A be an n × n matrix. Prove that the following statements are equivalent:
- A is invertible.
- Ax = b has exactly one solution for every n × 1 matrix b.
- Ax = b is consistent for every n × 1 matrix b.
- Ax = 0 has only the trivial solution x = 0.
The typical way to solve such an exercise is to show a series of implications. For instance, one can proceed by showing that (a) implies (b), that (b) implies (c), that (c) implies (d), and finally that (d) implies (a). These four implications show that the four statements are equivalent.
Another way would be to show that (a) is equivalent to (b) (by proving that (a) implies (b) and that (b) implies (a)), that (b) is equivalent to (c), and that (c) is equivalent to (d). However, this way requires proving six implications, which is clearly a lot more work than just proving four implications!
I have been given some similar tasks, and have already started proving some implications. Now I wonder, how many more implications do I have to prove? Can you help me determine this?
Input
On the first line one positive number: the number of testcases, at most 100. After that per testcase:
- One line containing two integers n (1 ≤ n ≤ 20000) and m (0 ≤ m ≤ 50000): the number of statements and the number of implications that have already been proved.
- m lines with two integers s1 and s2 (1 ≤ s1, s2 ≤ n and s1 ≠ s2) each, indicating that it has been proved that statement s1 implies statement s2.
Output
Per testcase:
- One line with the minimum number of additional implications that need to be proved in order to prove that all statements are equivalent.
Sample Input
2
4 0
3 2
1 2
1 3
Sample Output
4
2
解题心得：
- 题目说了一大堆废话，其实就是给你一个有向图，问你最少还需要添加多少条有向边可以将整个图变成强联通图。
- 其实想想就知道，可以先使用tarjan缩点，缩点之后会形成一个新的图，然后看图中出度为0和入度为0的点，因为这些点必然需要添一条边到图中，所以直接去取出度为0点的数目和入读为0的点的数目的最大值。为啥是最大值？很简单啊，将一条边添在出度为0的点和入度为0的点之间不就解决了两个点了吗，但是最后肯定要添加数目多的度为0的点啊。
- 注意一个坑点，那就如果可以直接缩为一个点那是不用添边的。
#include<stdio.h>
#include<iostream>
#include<cstring>
#include<stack>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn = 2e4+100;
vector <int> ve[maxn],maps[maxn],shrink[maxn];
bool vis[maxn];
int tot,num,indu[maxn],outdu[maxn],n,m,dfn[maxn],low[maxn],pre[maxn];
stack <int> st;
void init()//初始化很重要
{
while(!st.empty())
st.pop();
tot = num = 0;
memset(outdu,0,sizeof(outdu));
memset(indu,0,sizeof(indu));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(vis,0,sizeof(vis));
memset(pre,0,sizeof(pre));
for(int i=0;i<maxn;i++)
{
ve[i].clear();
shrink[i].clear();
maps[i].clear();
}
for(int i=0;i<m;i++)
{
int a,b;
scanf("%d%d",&a,&b);
ve[a].push_back(b);
}
}
void tarjan(int x)
{
dfn[x] = low[x] = ++tot;
vis[x] = true;
st.push(x);
for(int i=0;i<ve[x].size();i++)
{
int v = ve[x][i];
if(!dfn[v])
{
tarjan(v);
low[x] = min(low[x],low[v]);
}
else if(vis[v])
low[x] = min(low[x],dfn[v]);
}
if(low[x] == dfn[x])
{
while(1)
{
int now = st.top();
st.pop();
vis[now] = false;
shrink[num].push_back(now);
pre[now] = num;
if(now == x)
break;
}
num++;
}
}
void get_new_maps()
{
if(num == 1)
{
printf("0
");
return;
}
for(int i=1;i<=n;i++)
{
for(int j=0;j<ve[i].size();j++)
{
int a = i;
int b = ve[i][j];
if(pre[a] != pre[b])
{
outdu[pre[a]]++;
indu[pre[b]]++;
}
}
}
int sum_indu,sum_outdu;
sum_indu = sum_outdu = 0;
for(int i=0;i<num;i++)
{
if(!indu[i])
sum_indu++;
if(!outdu[i])
sum_outdu++;
}
printf("%d
",max(sum_indu,sum_outdu));
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
init();
for(int i=1;i<=n;i++)
{
if(!dfn[i])
tarjan(i);
}
get_new_maps();
}
return 0;
}