【BZOJ 1023】【SHOI 2008】cactus仙人掌图

良心的题解↓


http://z55250825.blog.163.com/blog/static/150230809201412793151890/

tarjan的时候如果是树边则做树形DP(遇到环就无视)，最后在tarjan回溯前扫一遍当前点为“最高点”的环，进行环上DP，这个环上DP是$O(n^2)$的，但如果我们用单调队列优化则是$O(n)$的

总复杂度$O(n)$真是无限仰膜OTZ

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 500003;
void read(int &k) {
k = 0; int fh = 1; char c = getchar();
for(; c < '0' || c > '9'; c = getchar())
if (c == '-') fh = -1;
for(; c >= '0' && c <= '9'; c = getchar())
k = (k << 1) + (k << 3) + c - '0';
k = k * fh;
}

struct node {int nxt, to;} E[N << 1];
int ans = 0, point
, n, m, cnt = 0, l, r;
int deep
, F
, fa
, DFN
, low
, Q[N << 1], tb[N << 1];

void ins(int x, int y) {E[++cnt] = (node) {point[x], y}; point[x] = cnt;}

void __(int rt, int x) {
int tot = deep[x] - deep[rt] + 1, num = tot << 1, top = tot >> 1;
for(int tmp = x; tmp != rt; tmp = fa[tmp]) tb[tot--] = F[tmp];
tb[1] = F[rt];
tot = deep[x] - deep[rt] + 1;
for(int i = 1; i <= tot; ++i) tb[tot + i] = tb[i];
l = r = 1; Q[1] = 1;
for(int i = 2; i <= num; ++i) {
while (l <= r && i - Q[l] > top) ++l;
ans = max(ans, tb[i] + i + tb[Q[l]] - Q[l]);
while (l <= r && tb[Q[r]] - Q[r] <= tb[i] - i) --r;
Q[++r] = i;
}
for(int i = 2; i <= tot; ++i)
F[rt] = max(F[rt], tb[i] + min(i - 1, tot - i + 1));
}

void _(int x) {
DFN[x] = low[x] = ++cnt;
for(int tmp = point[x]; tmp; tmp = E[tmp].nxt) {
int v = E[tmp].to;
if (v == fa[x]) continue;
if (!DFN[v]) {
fa[v] = x; deep[v] = deep[x] + 1;
_(v); low[x] = min(low[x], low[v]);
} else
low[x] = min(low[x], DFN[v]);
if (DFN[x] < low[v]) {
ans = max(ans, F[x] + F[v] + 1);
F[x] = max(F[x], F[v] + 1);
}
}
for(int tmp = point[x]; tmp; tmp = E[tmp].nxt) {
int v = E[tmp].to;
if (x == fa[v] || DFN[x] > DFN[v]) continue;
__(x, v);
}
}

int main() {
read(n); read(m);

int u, v, k;
for(int i = 1; i <= m; ++i) {
read(k); read(u);
for(--k; k; --k) {read(v); ins(u, v); ins(v, u); u = v;}
}

cnt = 0;
_(1);
printf("%d\n", ans);
return 0;
}

仙人掌虽然偏，但是不知道也不可以，这个代码是我磕了一晚上题解的成果QAQ