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

题目链接：

Description

The computer network of “Plunder & Flee Inc.” consists of n servers and m two-way communication links. Two servers can communicate either through a direct link, or through a chain of links, by relaying information from server to server. Current network setup enables communication for any pair of servers. The network administrator strives to maximize network reliability. Some communication links in the network were identified as critical. A failure on any critical link will split the network into disconnected segments. Company management responded to the administrator’s concerns and agreed to fund another communication link, provided that when the new link goes online the number of critical links will be minimized. Write a program that, given a network configuration, will pick a pair of servers to be connected by the new communication link. If several such pairs allow minimizing the number of critical links then any of them will be considered as a correct answer. Example. The following figure presents a network consisting of 7 servers and 7 communication links. Essential links are shown as bold lines. A new link connecting servers #1 and #7 (dotted line) can reduce the number of the critical links to only one

Input

The first line contains two space-delimited integer numbers n (the number of servers) and m (the number of communication links). The following m lines describe the communication links. Each line contains two space-delimited integers xi and yi, which define the IDs of servers connected by link number i. Servers are identified with natural numbers ranging from 1 to n.

Output

The output file should contain a single line with space-delimited integers x and y, the IDs of servers to be connected by the new link..

Sample Input

7 7 1 2 2 3 2 4 2 6 3 4 4 5 6 7

Sample Output

1 7

HINT

1 ≤ n ≤ 10 000; 1≤ m ≤ 100 000; 1 ≤ xi, yi ≤ n; xi ≠ yi.

题意：

给你一个无相图，加一条边，使得桥的个数最小化。

题解：

对原图缩点得到一颗树，求树的直径，把直径两端点连起来。

代码：

#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("
");
    //}
}

//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
", i,sccno[i]);
        if (sccno[i] == u) au = i + 1;
        if (sccno[i] == v) av = i + 1;
    }
    printf("%d %d
", au, av);
    return 0;
}