BZOJ 4016: [FJOI2014]最短路径树问题( 最短路 + 点分治 )

Posted 2020-06-13

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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