對樹的dfs序分塊，開啟了新世界的大門233
第一關鍵字是l所在的塊，第二關鍵字是r所在的塊，第三關鍵字是時間，分完塊後暴力莫隊即可
#include#include#include#includeusing namespace std;typedef long long ll;
const int n = 100003;
void read(int &k)
bool vis[n];
ll a[n], ans = 0;
struct time time[n];
struct edge e[n << 1];
struct node q[n];
int cal[n], n, m, q, v[n], w[n], f[n][17], deep[n], bel[n << 1];
int point[n], color[n], colorchange[n], pos[n << 1], l[n], r[n], cnt = 0;
void ins(int x, int y)
void _(int x, int fa)
for(int tmp = point[x]; tmp; tmp = e[tmp].nxt)
if (e[tmp].to != fa)
pos[r[x] = ++cnt] = x;
}int lca(int x, int y)
void update(int x)
else
vis[x] = !vis[x];
}void change(int a, int b)
else color[a] = b;
}void timechange(int &a, int b)
while (a > b)
}bool cmp(node a, node b)
int main()
for(int i = 1; i <= n; ++i) read(color[i]), colorchange[i] = color[i];
cnt = 0;
_(1, 0);
for(int i = 1; i <= q; ++i) ; colorchange[u] = v;}
else ;
else q[++tmp] = (node) ;
} }int nn = n << 1, sq = pow(nn, 2.0 / 3) * 0.5, sign = 0; cnt = 1;
for(int i = 1; i <= nn; ++i) }
sort(q + 1, q + tmp + 1, cmp);
int l = 1, r = 0, now = 0, tol, tor;
for(int i = 1; i <= tmp; ++i)
for(int i = 1; i <= tmp; ++i) printf("%lld\n", a[i]);
return 0;
}
dfs序分塊戰術核飛彈速度超快～
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