codeforces GYM 100114 J. Computer Network 无相图缩点+树的直径

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<stack>
using namespace std;

const int maxn = 10000 + 10;

struct Edge {
    int u,v,type;
    Edge(int u,int v,int type=0) : u(u),v(v),type(type) {}
};

vector<int> G[maxn];
vector<Edge> egs;
int n, m;

void addEdge(int u, int v) {
    egs.push_back(Edge(u,v));
    G[u].push_back(egs.size() - 1);
}

int pre[maxn], lowlink[maxn], sccno[maxn], dfs_clock, scc_cnt;
stack<int> S;

//无相图的边双联通分量 tarjan
void dfs(int u) {
    pre[u] = lowlink[u] = ++dfs_clock;
    S.push(u);
    for (int i = 0; i < G[u].size(); i++){
        Edge& e = egs[G[u][i]];
        if (e.type == 1) continue;
        egs[G[u][i] ^ 1].type ^= 1;//针对无相图，要标记反向边
        if (!pre[e.v]) {
            dfs(e.v);
            lowlink[u] = min(lowlink[u], lowlink[e.v]);
        }
        else if (!sccno[e.v]) {
            lowlink[u] = min(lowlink[u],pre[e.v]);
        }
    }
    if (lowlink[u] == pre[u]) {
        scc_cnt++;
        for (;;) {
            int x = S.top(); S.pop();
            sccno[x] = scc_cnt;
            if (x == u) break;
        }
    }
}

void find_scc() {
    dfs_clock = scc_cnt = 0;
    memset(sccno, 0, sizeof(sccno));
    memset(pre, 0, sizeof(pre));
    for (int i = 0; i < n; i++) if (!pre[i]) dfs(i);
}

vector<int> T[maxn];
//缩点得到的桥构成的树
void build_tree() {
    for (int i = 0; i < egs.size(); i++) {
        Edge& e = egs[i];
        if (e.type) continue;
        if (sccno[e.u] != sccno[e.v]) {
            T[sccno[e.u]].push_back(sccno[e.v]);
            T[sccno[e.v]].push_back(sccno[e.u]);
        }
    }
    //for (int i = 1; i <= scc_cnt; i++) {
    //    printf("%d:", i);
    //    for (int j = 0; j < T[i].size(); j++) {
    //        int v = T[i][j];
    //        printf("%d ", v);
    //    }
    //    printf("\n");
    //}
}

//求树的直径
void dfs2(int u,int fa,int d,int &dep,int &res) {
    if (dep < d) dep = d, res = u;
    for (int i = 0; i < T[u].size(); i++) {
        int v = T[u][i];
        if (v == fa) continue;
        dfs2(v, u, d + 1, dep, res);
    }
}

void init() {
    for (int i = 0; i < n; i++) G[i].clear(), T[i].clear();
    T[n].clear();
}

int main() {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    scanf("%d%d", &n, &m);
    init();
    for (int i = 0; i < m; i++) {
        int u, v;
        scanf("%d%d", &u, &v),u--,v--;
        addEdge(u, v);
        addEdge(v, u);
    }
    find_scc();
    build_tree();
    int u, v, dep;
    dep=-1,dfs2(1, -1, 0, dep, u);
    dep=-1,dfs2(u, -1, 0, dep, v);
    int au, av;
    for (int i = 0; i < n; i++) {
        //printf("sccno[%d]:%d\n", i,sccno[i]);
        if (sccno[i] == u) au = i + 1;
        if (sccno[i] == v) av = i + 1;
    }
    printf("%d %d\n", au, av);
    return 0;
}