2014
03-06

# Dominos

Dominos are lots of fun. Children like to stand the tiles on their side in long lines. When one domino falls, it knocks down the next one, which knocks down the one after that, all the way down the line. However, sometimes a domino fails to knock the next one down. In that case, we have to knock it down by hand to get the dominos falling again.

Your task is to determine, given the layout of some domino tiles, the minimum number of dominos that must be knocked down by hand in order for all of the dominos to fall.

The first line of input contains one integer specifying the number of test cases to follow. Each test case begins with a line containing two integers, each no larger than 100 000. The first integer n is the number of domino tiles and the second integer m is the number of lines to follow in the test case. The domino tiles are numbered from 1 to n. Each of the following lines contains two integers x and y indicating that if domino number x falls, it will cause domino number y to fall as well.

The first line of input contains one integer specifying the number of test cases to follow. Each test case begins with a line containing two integers, each no larger than 100 000. The first integer n is the number of domino tiles and the second integer m is the number of lines to follow in the test case. The domino tiles are numbered from 1 to n. Each of the following lines contains two integers x and y indicating that if domino number x falls, it will cause domino number y to fall as well.

1
3 2
1 2
2 3

1

//IN THE NAME OF GOD
#include <iostream>
#include <vector>
#include <stack>
#include <string>
#include <algorithm>
#include <cmath>
#include <set>
#include <queue>
#include <map>
#include <fstream>
#include <utility>
#include <sstream>
#include <list>
#include <iomanip>
#include <functional>
#include <deque>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <complex>
#include <climits>
#include <cctype>
#include <cassert>
#include <bitset>
#include <limits>
#include <numeric>

using namespace std;

#define timestamp(x) printf("Time : %.3lf s.\n", clock()*1.0/CLOCKS_PER_SEC)
#define INF 1000000000
#define pii pair < int , int >
#define MP make_pair
#define MOD 1000000007
#define EPS 1e-9
#define LL long long
#define MAXN 200000+10

int n, m;
vector < int > dfs_low, dfs_num, visited, S , numgroup , tmp ;
int dfsNumberCounter, numSCC, Z;
vector < vector < int > > group;

void tarjanSCC(int u) {
dfs_low[u] = dfs_num[u] = dfsNumberCounter++; // dfs_low[u] <= dfs_num[u]
S.push_back(u); // stores u in a vector based on order of visitation
visited[u] = 1;
for (int j = 0; j < (int)AdjList[u].size(); j++) {
if (dfs_num[v] == 0 )
tarjanSCC(v);
if (visited[v]) // condition for update
dfs_low[u] = min(dfs_low[u], dfs_low[v]);
}
if (dfs_low[u] == dfs_num[u]) { // if this is a root (start) of an SCC
//printf("SCC %d:", ++numSCC); // this part is done after recursion
Z++;
tmp.clear();
while (1) {
int v = S.back(); S.pop_back(); visited[v] = 0;
tmp.push_back(v);
numgroup[v] = Z;
if (u == v) break;
}
group.push_back(tmp);
//printf("\n");
}
}

void init(){
dfs_low.clear(), dfs_num.clear(), visited.clear(), S.clear() , numgroup.clear() , group.clear() ;
for (int i = 0; i < MAXN; i++) AdjList[i].clear() , RAdjList[i].clear() ;
numSCC = 0;
dfs_low.assign(MAXN, 0), dfs_num.assign(MAXN, 0), visited.assign(MAXN, 0), numgroup.assign(MAXN, 0);
dfsNumberCounter = numSCC = 0;
Z = 0;
}

int main()
{
ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen("i.txt", "r", stdin);
//freopen("o.txt", "w", stdout);
#endif
int t, u, v;
cin >> t;
while (t--){
cin >> n >> m;
init();
for (int i = 0; i < m; i++){
cin >> u >> v;
}
for (int i = 1 ; i <= n; i++){
if (dfs_num[i] == 0){
tarjanSCC(i);
}
}
int f;
for (int i = 0; i < group.size(); i++){
f = 0;
for (int j = 0; j < group[i].size() && !f ; j++){
v = group[i][j];
for (int t = 0; t < RAdjList[v].size() && !f ; t++){
}