CSU 1612: Destroy Tunnels strongly connected component Kosaraju algorithm, tunnelskosaraju

CSU 1612: Destroy Tunnels strongly connected component Kosaraju algorithm, tunnelskosaraju
Link: zookeeper

Http://acm.csu.edu.cn/OnlineJudge/problem.php? Id = 1612

Give A matrix A the size of N * N, B = A ^ 1 + A ^ 2 + A ^ 3 + .... whether non-zero items exist in A ^ n and B.

The question can be converted to N vertices numbered 1-N. For any vertices, the question goes through any step to reach u (u is any of all vertices ). In Discrete Mathematics, matrix-related knowledge A ^ k represents the number of methods in the matrix from I to j in step k.

The question is to determine whether the entire graph is strongly connected.

After learning the Kosaraju algorithm, you can easily solve the problem. For details, see:

Http://www.cnblogs.com/tnt_vampire/archive/2010/04/12/1709895.html

Http://blog.csdn.net/dm_vincent/article/details/8554244

#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <sstream>#include <cstdio>#include <vector>#include <cmath>#include <queue>#include <stack>#include <set>#include <map>#define lson o<<1, l, m#define rson o<<1|1, m+1, r#define PII pair<int, int>#define ALL(x) x.begin(),x.end()#define mem(a) memset(a,0,sizeof(a))typedef long long ll;const double pi = acos(-1.0);const int MAX = 0x3f3f3f3f;const ll mod = 1000000007ll;const int N = 1005;using namespace std; int T, n, x;int a[N][N], scc[N], cnt, v[N];vector <int> G[N], G2[N], S; void dfs1(int u) {    if(v[u]) return;    v[u] = 1;    for(int i = 0; i < G[u].size(); i++) {        dfs1(G[u][i]);    }    S.push_back(u);} void dfs2(int u) {    if(scc[u]) return;    scc[u] = cnt;    for(int i = 0; i < G2[u].size(); i++) {        dfs2(G2[u][i]);    }} int main() {     //  freopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);    cin >> T;    while(T--) {        cin >> n;        cnt = 0;        mem(scc);        mem(v);        for(int i = 0; i < n; i++) {            G2[i].clear();            G[i].clear();        }                 S.clear();        for(int i = 0; i < n; i++) {            for(int j = 0; j < n; j++) {                scanf("%d", &x);                if(x) {                    G[i].push_back(j);                    G2[j].push_back(i);                }            }        }                          for(int i = 0; i < n; i++) {            dfs1(i);        }        for(int i = n-1; i >= 0; i--) {            if(!scc[S[i]]) {                cnt++;                dfs2(S[i]);            }        }        puts(cnt == 1 ? "not exists" : "exists");                      }    return 0;     }