Bzoj 1023 cactus chart

Bzoj 1023 cactus chart
Description calculate the diameter of a cactus image Solution the cactus image has a personality quality, one side is either cut side or in the ring, then we can conduct it Dp spicy! Ling F [u] indicates the longest chain length of the subtree rooted in u. If U? V If the bridge is used, the transfer is Ans = max (ans, f [u] + f [v] + 1), f [u] = max (f [u], f [v] + 1) Because at present, f [u] is updated by its children. If it is a ring, it is changed to a chain, and the ans are output from the monotonous queue dp, then, use the f value on the ring to update the f [u] value. For specific implementation, see Code.

#include 
  
   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\n", ans);    return 0;}