Description
求1个神仙掌图的直径
Solution
神仙掌图有个性质，1条边要末是割边要末就是在环内，那末我们可以对它进行Dp辣！
令f[u]表示以u为根的子树最长链长度
如果u?v是桥的话转移就是ans=max(ans,f[u]+f[v]+1)，f[u]=max(f[u],f[v]+1)，由于当前f[u]都是由它的孩子更新来的
如果是环的话，变环为链，用单调队列dp出ans,然后用环上的f值更新f[u]的值就能够了，具体实现见代码
Code
using namespace std;
const int N = 100005,M = N << 1;
int ans,ind,tot,cnt,fa[N],cir[N << 1],to[M << 1],nxt[M << 1],head[N],dfn[N],low[N],f[N];
inline int read(int &t) {
int f = 1;char c;
while (c = getchar(),c < '0' || c > '9') if (c == '-') f = -1;
t = c - '0';
while (c = getchar(),c >= '0' && c <= '9') t = t * 10 + c - '0';
t *= f;
}
struct data {
int p,w;
}q[N];
void add(int u,int v) {
to[tot] = v,nxt[tot] = head[u],head[u] = tot++;
to[tot] = u,nxt[tot] = head[v],head[v] = tot++;
}
void gao() {
int h = 1,r = 1;
for (int i = 1; i <= cnt; ++i) cir[cnt + i] = cir[i];
for (int i = 1; i <= (cnt << 1); ++i) {
while (h < r && i - q[h].p > cnt / 2) ++h;
while (h < r && q[r].w <= f[cir[i]] - i) --r;
q[++r].p = i,q[r].w = f[cir[i]] - i;
ans = max(ans,f[cir[i]] + i + q[h].w);
}
}
void dfs(int u) {
low[u] = dfn[u] = ++ind;
for (int i = head[u],v; ~i; i = nxt[i]) {
v = to[i];
if (fa[v] != 0 && v != fa[u]) low[u] = min(low[u],dfn[v]);
if (fa[v] == 0) {
fa[v] = u;
dfs(v);
low[u] = min(low[u],low[v]);
}
}
for (int i = head[u],v; ~i; i = nxt[i]) {
v = to[i];
if (fa[v] == u && low[v] > dfn[u]) { //bridge
ans = max(ans,f[u] + f[v] + 1);
f[u] = max(f[u],f[v] + 1);
}
if (fa[v] != u && dfn[u] < dfn[v]) { //circle
cnt = 0;
while (v != fa[u]) cir[++cnt] = v,v = fa[v];
gao();
for (int j = 1; j < cnt; ++j) f[u] = max(f[u],f[cir[j]] + min(j,cnt - j));
}
}
}
int main() {
int n,m;
memset(head,-1,sizeof(head));
read(n),read(m);
for (int i = 1,x,y,z; i <= m; ++i) {
read(x),read(y);
for (int j = 1; j < x; ++j) {
read(z);
add(y,z);
y = z;
}
}
fa[1] = -1;
dfs(1);
printf("%d
",ans);
return 0;
}
