ACM Arabella Collegiate Programming Contest 2015 H. Capital City 边连通分量

#include <bits/stdc++.h>

using namespace std;

const int maxn = 100005;
const long long inf = 1e15 + 5;

struct Edge
{
long long u, v, c;
} e[maxn<<2];
struct Ed
{
long long v, c;
};

int n, m, low[maxn], pre[maxn], tim, ebcc_cnt, du[maxn];
long long k, len, dis[maxn][2];
vector<int> G[maxn];
vector<Ed> ed[maxn];
int isbri[maxn<<4];
bool vis[maxn];

void init()
{
ebcc_cnt = tim = 0;
for (int i = 1; i <= n; i++) G[i].clear();
memset(isbri, 0, sizeof(isbri));
memset(pre, 0, sizeof(pre));
memset(du, 0, sizeof(du));
}

void tarjan(int u, int fa)
{
low[u] = pre[u] = ++tim;
for (int i = 0; i < G[u].size(); i++)
{
int tmp = G[u][i];
int v = e[tmp].v;
if (!pre[v])
{
tarjan(v, u);
low[u] = min(low[u], low[v]);
if (low[v] > pre[u])
isbri[tmp] = isbri[tmp^1] = true; //标记为桥
}
else if (fa != v)
low[u] = min(low[u], pre[v]);
}
}

void dfs(int u)
{
pre[u] = ebcc_cnt;
for (int i = 0; i < G[u].size(); i++)
{
int tmp = G[u][i];
if (isbri[tmp]) continue;
int v = e[tmp].v;
if (pre[v]) continue;
dfs(v);
}
}

void find_ebcc()
{
tarjan(1, -1);
memset(pre, 0, sizeof(pre));
for (int i = 1; i <= n; i++)
{
if (!pre[i])
{
ebcc_cnt++;
dfs(i);
}
}
}

void BFS(int s, int ca)
{
memset(vis, 0, sizeof(vis));
queue<Ed> q;
q.push((Ed)
{
s, 0
});
vis[s] = 1;
while (q.size())
{
Ed tmp = q.front();
q.pop();
dis[tmp.v][ca] = tmp.c;
for (int i = 0; i < ed[tmp.v].size(); i++)
{
Ed xx = ed[tmp.v][i];
if (!vis[xx.v])
{
vis[xx.v] = 1;
q.push((Ed)
{
xx.v, xx.c + tmp.c
});
}
}
}
}

void dfs_len(int x, int fa, long long dep)
{
if (dep > len)
{
k = x;
len = dep;
}
for (int i = 0; i < ed[x].size(); i++)
{
Ed tmp = ed[x][i];
if (tmp.v == fa) continue;
dfs_len(tmp.v, x, dep + tmp.c);
}
}

int main()
{
int t = 1;
cin >> t;

while (t--)
{
cin >> n >> m;
init();
for (int i = 1; i <= m; i++)
{
int u, v, c;
scanf("%d%d%d", &u, &v, &c);
e[i<<1|1].u = u, e[i<<1|1].v = v, e[i<<1|1].c = c;
e[i<<1].u = v, e[i<<1].v = u, e[i<<1].c = c;
G[u].push_back(i<<1|1);
G[v].push_back(i<<1);
}
find_ebcc();
int tot = m<<1|1;
for (int i = 1; i <= ebcc_cnt; i++) ed[i].clear();
for (int i = 3; i <= tot; i += 2)
{
if (isbri[i])
{
int u = e[i].v, v = e[i].u;
ed[pre[u]].push_back((Ed)
{
pre[v], e[i].c
});
ed[pre[v]].push_back((Ed)
{
pre[u], e[i].c
});
}
}
len = -1;
dfs_len(1, -1, 0);
int st = k;
len = -1;
dfs_len(st, -1, 0);
BFS(st, 0);
BFS(k, 1);
long long inx = n + 1, dd = inf;
for (int i = 1; i <= n; i++)
{
int pr = pre[i];
if (dis[pr][0] + dis[pr][1] != len) continue;
long long tmp = max(dis[pr][0], dis[pr][1]);
if (tmp < dd)
{
inx = i;
dd = tmp;
}
}
printf("%lld %lld\n",inx,dd);
}
return 0;
}