Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P4768 [NOI2018] 归程

P4768 [NOI2018] 归程

求出 Kruskal 重构树,那么对于每个询问,找到符合条件的深度最小的点,那么这个点子树内的所有点都是可以直接到达的,对于子树中的每一个点预处理出到 $1$ 号点的最短路,那么答案即为子树内最短路的最小值。

查看代码
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#include <cstdio>
#include <vector>
#include <queue>
#include <algorithm>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 1) + (x << 3) + c - '0';
flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar('0' + x % 10);
}
typedef pair <int, int> PII;
const int N = 4e5 + 10, M = 20, inf = 2e9;
int n, m, q, d[N], p[N], w[N], f[N][M], v[N][M];
vector <int> g[N];
struct Edge
{
int u, v, w;
bool operator < (const Edge &_) const
{ return w > _.w; }
} e[N];
namespace SP
{
const int M = 8e5 + 10;
bool vis[N];
int idx, hd[N], nxt[M], edg[M], wt[M];
void add (int a, int b, int c)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
wt[idx] = c;
}
void init ()
{
idx = 0;
for (int i = 1; i <= n; ++i)
hd[i] = -1, vis[i] = false;
}
void solve ()
{
d[1] = 0;
for (int i = 2; i <= n; ++i) d[i] = inf;
priority_queue <PII, vector <PII>, greater<PII> > q;
q.push(mp(0, 1));
while (!q.empty())
{
int t = q.top().se; q.pop();
if (vis[t]) continue;
vis[t] = true;
for (int i = hd[t]; ~i; i = nxt[i])
if (wt[i] < d[edg[i]] - d[t])
{
d[edg[i]] = d[t] + wt[i];
q.push(mp(d[edg[i]], edg[i]));
}
}
}
}
int fa (int x) { return p[x] == x ? x : p[x] = fa(p[x]); }
void dfs (int x)
{
if (!g[x].empty()) d[x] = inf;
for (int i : g[x])
{
f[i][0] = x, v[i][0] = w[x];
for (int k = 0; k + 1 < M; ++k)
{
f[i][k + 1] = f[f[i][k]][k];
v[i][k + 1] = min(v[i][k], v[f[i][k]][k]);
}
dfs(i);
d[x] = min(d[x], d[i]);
}
}
int main ()
{
int T; read(T);
while (T--)
{
read(n, m);
for (int i = 1; i <= n; ++i)
g[i].clear(), p[i] = i;
SP::init();
for (int i = 1, a; i <= m; ++i)
{
read(e[i].u, e[i].v, a, e[i].w);
SP::add(e[i].u, e[i].v, a);
SP::add(e[i].v, e[i].u, a);
}
SP::solve();
sort(e + 1, e + m + 1);
int tot = n;
for (int i = 1; i <= m; ++i)
{
int a = fa(e[i].u), b = fa(e[i].v);
if (a == b) continue;
p[a] = p[b] = ++tot, p[tot] = tot;
g[tot].clear();
w[tot] = e[i].w;
g[tot].pb(a), g[tot].pb(b);
}
int rt = fa(1); f[rt][0] = 0, dfs(rt);
int S;
bool op;
read(q, op, S);
for (int u, p, last = 0; q; --q)
{
read(u, p);
u = (u + op * last - 1) % n + 1, p = (p + op * last) % (S + 1);
for (int i = M - 1; ~i; --i)
if (f[u][i] && v[u][i] > p) u = f[u][i];
write(last = d[u]), puts("");
}
}
return 0;
}