uvaLive_4287_Proving Equivalences(缩点)

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 statements1 implies statements2.

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

题意：就是给你一个有向图，问你至少添加多少条边，使得该图为强连通图。

分析：刚开始学tarjan，做这道题的时候，只想到了用tarjan来缩点，思路也就在这里戛然而止了。后来是问了鑫神才知道怎么处理的。只需要统计出入度为0的点总和sum_in以及出度为0的点的总和sum_out，最后取其最大值max(sum_in,sum_out)即可。

而为什么这样子做就可以呢？我觉得是，首先强连通图一定存在一个回路使得它包含所有的点，当我们这样取了之后，所有点的的出度和入度一定都大于1，也就是该图一定存在一个回路，但是存在回路不一定包含所有的点，而我们却可以适当地连边构图使得所有点一定都在这个回路上。

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=10294

代码清单：

#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<ctime>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;

const int maxn = 20000 + 5;
const int maxv = 50000 + 5;

int T,n,m,s1,s2;
int dfn[maxn];        //深度优先访问次序
int low[maxn];        //能追溯到的最早次序
vector<int>G[maxn];   //存图
stack<int>sta;        //存储已遍历的点
bool InStack[maxn];   //是否在栈中
int index1;           //索引号
int index2;           //强连通分量个数
int sccno[maxn];      //缩点
int indegree[maxn];   //入度
int outdegree[maxn];  //出度

void init(){
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(sccno,0,sizeof(sccno));
    memset(InStack,false,sizeof(InStack));
    memset(indegree,0,sizeof(indegree));
    memset(outdegree,0,sizeof(outdegree));
    while(!sta.empty()) sta.pop();
    for(int i=1;i<=n;i++) G[i].clear();
    index1=0;
    index2=0;
}

void input(){
    scanf("%d%d",&n,&m);
    for(int i=0;i<m;i++){
        scanf("%d%d",&s1,&s2);
        G[s1].push_back(s2);
    }
}

void tarjan(int u){
    dfn[u]=low[u]=++index1;
    sta.push(u);
    InStack[u]=true;
    for(int i=0;i<G[u].size();i++){
        int v=G[u][i];
        if(!dfn[v]){
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        if(InStack[v]){
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(low[u]==dfn[u]){
        index2++;
        while(!sta.empty()){
            int j=sta.top();
            sta.pop();
            InStack[j]=false;
            sccno[j]=index2;
            if(j==u) break;
        }
    }
}

void findscc(){
    for(int i=1;i<=n;i++){
        if(!dfn[i]) tarjan(i);
    }
}

int work(){
    findscc();
    if(index2==1) return 0;
    for(int u=1;u<=n;u++){
        for(int i=0;i<G[u].size();i++){
            int v=G[u][i];
            if(sccno[u]!=sccno[v]){
                int uu=sccno[u];
                int vv=sccno[v];
                outdegree[uu]++;
                indegree[vv]++;
            }
        }
    }
    int in=0,out=0;
    for(int i=1;i<=index2;i++){
        if(indegree[i]==0) in++;
        if(outdegree[i]==0) out++;
    }
    return max(in,out);
}

void solve(){
    printf("%d\n",work());
}

int main(){
    scanf("%d",&T);
    while(T--){
        init();
        input();
        solve();
    }return 0;
}