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先跑出最短路的图, 然后对于每个点按照序号从小到大访问孩子, 就可以搞出符合题目的树了. 然后就是经典的点分治做法了. 时间复杂度O(M log N + N log N)
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#include<queue>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 30009;
const int INF = 0X3F3F3F3F;
inline int getint() {
char c = getchar();
for(; !isdigit(c); c = getchar());
int ret = 0;
for(; isdigit(c); c = getchar())
ret = ret * 10 + c - ‘0‘;
return ret;
}
int N, K, n = 0;
int d[maxn], seq[maxn << 1], w[maxn << 1], r[maxn << 1], L[maxn], R[maxn];
bool vis[maxn];
inline void Max(int &x, int t) {
if(t > x) x = t;
}
struct edge {
int t, w;
bool f;
edge* n;
} E[maxn << 2], *pt = E, *H[maxn];
inline void AddEdge(int u, int v, int w) {
pt->f = 0, pt->t = v, pt->w = w, pt->n = H[u], H[u] = pt++;
}
struct node {
int n, w;
node(int _n, int _w) : n(_n), w(_w) {
}
bool operator < (const node &o) const {
return w > o.w;
}
};
priority_queue<node> q;
void Dijkstra() {
for(int i = 0; i < N; i++) d[i] = INF;
d[0] = 0;
q.push(node(0, 0));
while(!q.empty()) {
node o = q.top(); q.pop();
if(d[o.n] != o.w) continue;
for(edge* e = H[o.n]; e; e = e->n) if(d[e->t] > d[o.n] + e->w) {
d[e->t] = d[o.n] + e->w;
q.push(node(e->t, d[e->t]));
}
}
}
void Init() {
N = getint();
int m = getint();
K = getint();
while(m--) {
int u = getint() - 1, v = getint() - 1, w = getint();
AddEdge(u, v, w), AddEdge(v, u, w);
}
}
namespace F {
edge E[maxn << 1], *pt = E, *H[maxn];
bool vis[maxn];
int L[maxn], cnt[maxn], sz[maxn];
int Root, mn, n, ans, tot;
inline void AddEdge(int u, int v, int w) {
pt->t = v, pt->w = w, pt->n = H[u], H[u] = pt++;
}
void DFS(int x, int fa = -1) {
int mx = 0;
sz[x] = 1;
for(edge* e = H[x]; e; e = e->n) if(e->t != fa && !vis[e->t]) {
DFS(e->t, x);
sz[x] += sz[e->t];
Max(mx, sz[e->t]);
}
Max(mx, n - sz[x]);
if(mx < mn)
mn = mx, Root = x;
}
void dfs_upd(int x, int fa, int d, int c) {
if(c >= K) return;
if(cnt[K - c - 1] && d + L[K - c - 1] > ans)
ans = d + L[K - c - 1], tot = cnt[K - c - 1];
else if(d + L[K - c - 1] == ans)
tot += cnt[K - c - 1];
c++;
for(edge* e = H[x]; e; e = e->n)
if(e->t != fa && !vis[e->t]) dfs_upd(e->t, x, d + e->w, c);
}
void dfs_add(int x, int fa, int d, int c) {
if(c >= K) return;
if(d > L[c])
L[c] = d, cnt[c] = 1;
else if(d == L[c])
cnt[c]++;
c++;
for(edge* e = H[x]; e; e = e->n)
if(e->t != fa && !vis[e->t]) dfs_add(e->t, x, d + e->w, c);
}
void Solve(int x) {
mn = maxn, DFS(x);
for(int i = 0; i <= sz[x]; i++) L[i] = cnt[i] = 0;
x = Root;
cnt[0] = 1;
for(edge* e = H[x]; e; e = e->n) if(!vis[e->t]) {
dfs_upd(e->t, x, e->w, 1);
dfs_add(e->t, x, e->w, 1);
}
DFS(x);
vis[x] = true;
for(edge* e = H[x]; e; e = e->n)
if(!vis[e->t]) n = sz[e->t], Solve(e->t);
}
void Work() {
for(int i = 0; i < N; i++)
vis[i] = false;
ans = tot = 0;
n = N;
Solve(0);
printf("%d %d\n", ans, tot);
}
}
bool Cmp(const int &l, const int &r) {
return seq[l] < seq[r];
}
void dfs(int x) {
seq[L[x] = ++n] = x;
r[n] = n;
vis[x] = true;
for(edge* e = H[x]; e; e = e->n) if(!vis[e->t] && d[e->t] == d[x] + e->w) {
seq[++n] = e->t;
w[n] = e->w;
r[n] = n;
}
R[x] = n;
sort(r + L[x], r + R[x] + 1, Cmp);
for(int i = L[x]; i <= R[x]; i++) if(!vis[seq[r[i]]]) {
F::AddEdge(x, seq[r[i]], w[r[i]]);
F::AddEdge(seq[r[i]], x, w[r[i]]);
dfs(seq[r[i]]);
}
}
void Build() {
for(int i = 0; i < N; i++) vis[i] = false;
dfs(0);
}
int main() {
Init();
Dijkstra();
Build();
F::Work();
return 0;
}
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BZOJ 4016: [FJOI2014]最短路径树问题( 最短路 + 点分治 )
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原文地址：http://www.cnblogs.com/JSZX11556/p/5179125.html
