hdu2767Proving Equivalences(强连通+缩点)

Proving Equivalences

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1962    Accepted Submission(s): 743

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:

1. A is invertible.
2. Ax = b has exactly one solution for every n × 1 matrix b.
3. Ax = b is consistent for every n × 1 matrix b.
4. 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
   
   
   
   

 

Source

NWERC 2008

 

Recommend

lcy

题目大意：给一张有向图，求最少加几条边使这个图强连通。

题目分析：先求这张图的强连通分量，如果为1，则输出0，不需要加边已经是强连通的了。否则缩点。遍历原图的所有边，如果2个点在不同的强连通分量里面，建边。构成一张新图。统计新图中点的入度和出度，取入度=0和出度=0的最大值。

因为求强连通缩点后，整张图就变成了一个DAG，要使之强连通，只需要将入度=0和出度=0的点加边即可，要保证加边后没有入度和出度为0的点，所以取两者最大值。

求强连通的话这三种算法任选一种，我用的Gabow算法。

详情请见代码：

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 20001;
const int M = 50001;
struct edge
{
    int to,next;
}g[M];
int head[N];
int stack1[N];
int stack2[N];
int scc[N];
int vis[N];
int in[N];
int out[N];
int n,m;

void init()
{
    int i;
    for(i = 1;i <= n;i ++)
    {
        head[i] = -1;
        scc[i] = vis[i] = in[i] = out[i] = 0;
    }
}

void dfs(int cur,int &sig,int &num)
{
    vis[cur] = ++sig;
    stack1[++stack1[0]] = cur;
    stack2[++stack2[0]] = cur;
    for(int i = head[cur];i != -1;i = g[i].next)
    {
        if(vis[g[i].to] == 0)//不在栈中，即未访问过
            dfs(g[i].to,sig,num);
        else
            if(scc[g[i].to] == 0)//在栈中，已访问过，又访问到，所以出现了环，删除环
            {
                while(vis[stack2[stack2[0]]] > vis[g[i].to])
                    stack2[0] --;
            }
    }
    if(stack2[stack2[0]] == cur)
    {
        ++num;
        stack2[0] --;
        do
        {
            scc[stack1[stack1[0]]] = num;
        }while(stack1[stack1[0] --] != cur);
    }
}

int Gabow()
{
    int i,sig,ret;
    //memset(vis,0,sizeof(vis));
    //memset(scc,0,sizeof(scc));
    stack1[0] = stack2[0] = sig = ret = 0;
    for(i = 1;i <= n;i ++)
        if(!vis[i])
            dfs(i,sig,ret);
    return ret;
}

int solve()
{
    int i,j;
    int num = Gabow();
    if(num == 1)
        return 0;
    //printf("num:%d\n",num);
    //memset(in,0,sizeof(in));
    //memset(out,0,sizeof(out));
    for(i = 1;i <= n;i ++)
    {
        for(j = head[i];j != -1;j = g[j].next)
        {
            if(scc[i] != scc[g[j].to])
            {
                out[scc[i]] ++;
                in[scc[g[j].to]] ++;
            }
        }
    }
    int cnt1,cnt2;
    cnt1 = cnt2 = 0;
    for(i = 1;i <= num;i ++)
    {
        if(in[i] == 0)
            cnt1 ++;
        if(out[i] == 0)
            cnt2 ++;
    }
    return cnt1 > cnt2?cnt1:cnt2;
}
int nextint()
{
    char c;
    int ret;
    while((c = getchar()) > '9' || c < '0')
        ;
    ret = c - '0';
    while((c = getchar()) >= '0' && c <= '9')
        ret = ret * 10 + c - '0';
    return ret;
}
void print(int x)
{
    if(!x)
        return;
    print(x/10);
    putchar(x%10 + '0');
}

int main()
{
    int i,j,a,b,t;
    //scanf("%d",&t);
    t = nextint();
    while(t --)
    {
        //scanf("%d%d",&n,&m);
        n = nextint();
        m = nextint();
        init();
        //memset(head,-1,sizeof(head));
        for(i = 1;i <= m;i ++)
        {
            //scanf("%d%d",&a,&b);
            a = nextint();
            b = nextint();
            g[i].to = b;
            g[i].next = head[a];
            head[a] = i;
        }
        //printf("%d\n",solve());
        int ans = solve();
        if(ans == 0)
            putchar('0');
        else
            print(ans);
        putchar(10);
    }
    return 0;
}
//46MS	1700K
//125MS	1708K