Blog of RuSun

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

K 短路模板

可持久化可并堆解决 K 短路问题。

对反向图建出最短路径树,将非树边的权值改为替换为这个边的变化值,转化问题:在非树边上选择若干路径,满足要么要相邻的,要么是树上的祖孙关系。对于每个点开始的可以替换的边建一个堆,注意到后者需要将一个堆再合并到儿子上,暴力做时空复杂度爆炸,所以用可持久化可并堆。最终复杂度 $O(m \log m + k \log k)$ 。

一个坑点,从终点的出边一定要删掉,否则会给终点多加一个堆中的点。

查看代码
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#include <cstdio>
#include <queue>
#include <vector>
#define pb push_back
#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 = 5e3 + 10, M = 2e5 + 10, T = 1e7 + 10, inf = 1e9;
vector <int> e[N];
bool vis[N];
int n, m, K, st, ed, d[N], pre[N];
int idx, hd[N], nxt[M << 1], edg[M << 1], wt[M << 1];
priority_queue <PII, vector <PII>, greater<PII> > q;
struct Node
{
int l, r, w, d, ed;
void init (int _w, int _ed)
{ w = _w, ed = _ed, d = 1; }
} tr[T];
int tot, rt[N];
int merge (int x, int y)
{
if (!x || !y) return x | y;
if (tr[x].w > tr[y].w) swap(x, y);
tr[++tot] = tr[x], x = tot;
tr[x].r = merge(tr[x].r, y);
if (tr[tr[x].r].d > tr[tr[x].l].d) swap(tr[x].l, tr[x].r);
tr[x].d = tr[tr[x].r].d + 1;
return x;
}
void dfs (int u)
{
for (int v : e[u])
rt[v] = merge(rt[v], rt[u]), dfs(v);
}
void add (int a, int b, int c)
{ nxt[idx] = hd[a], hd[a] = idx, edg[idx] = b, wt[idx] = c, ++idx; }
int main ()
{
read(n, m, K, st, ed);
for (int i = 1; i <= n; ++i) hd[i] = -1;
for (int a, b, c; m; --m)
{
read(a, b, c);
if (a ^ ed) add(a, b, c), add(b, a, c);
}
for (int i = 1; i <= n; ++i) d[i] = inf;
d[ed] = 0;
q.push({d[ed], ed});
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 (i & 1)
if (d[t] + wt[i] < d[edg[i]])
{
pre[edg[i]] = i;
d[edg[i]] = d[t] + wt[i];
q.push({d[edg[i]], edg[i]});
}
}
for (int i = 1; i <= n; ++i)
if (i ^ ed) e[edg[pre[i] ^ 1]].pb(i);
for (int i = 1; i < idx; i += 2) if (pre[edg[i]] ^ i)
{
tr[++tot].init(wt[i] + d[edg[i ^ 1]] - d[edg[i]], edg[i ^ 1]);
rt[edg[i]] = merge(rt[edg[i]], tot);
}
dfs(ed);
if (rt[st]) q.push({tr[rt[st]].w, rt[st]});
while (K--)
{
if (q.empty()) return puts("-1"), 0;
int k = q.top().fi, t = q.top().se; q.pop();
write(k + d[st]), puts("");
if (tr[t].l) q.push({tr[tr[t].l].w + k - tr[t].w, tr[t].l});
if (tr[t].r) q.push({tr[tr[t].r].w + k - tr[t].w, tr[t].r});
if (rt[tr[t].ed]) q.push({k + tr[rt[tr[t].ed]].w, rt[tr[t].ed]});
}
return 0;
}