「Bzoj 1999」「Noip2007」树网的核加强版 (树的直径+尺取法)

BZOJ 1999
题意:在直径上找到一个区间使得这个区间上所有点的偏心距最小。

容易想到用单调队列CF 1004E的做法。

这里其实偏心距求个最大值即可,就不用单调队列了,直接尺取。

突然发现这题和CF 1004E是一题。。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<set>
#include<queue>
#define ms(i, j) memset(i, j, sizeof i)
#define LL long long
#define db double
#define fir first
#define sec second
#define mp make_pair
using namespace std;
namespace flyinthesky {
const int MAXN = 500000 + 5;
struct edge {
int v, w, nxt;
} ed[MAXN * 2];
int n, s, en, hd[MAXN], whw1, whw2, vis[MAXN], cnt;
LL dis[MAXN];
void ins(int x, int y, int w) {ed[++en] = (edge){y, w, hd[x]}, hd[x] = en;}
void dfs1(int u, int fa) {
for (int i = hd[u]; i > 0; i = ed[i].nxt) {
edge &e = ed[i];
if (e.v != fa) {
dis[e.v] = dis[u] + e.w;
dfs1(e.v, u);
}
}
}
int dfs2(int u, int fa) {
if (u == whw2) return vis[u] = 1, true;
int fl = 0;
for (int i = hd[u]; i > 0; i = ed[i].nxt) {
edge &e = ed[i];
if (e.v != fa) {
fl |= dfs2(e.v, u);
if (fl) vis[u] = 1;
}
}
return fl;
}
LL gg = 0ll;
void dfs4(int u, LL D, int fa) {
gg = max(gg, D);
for (int i = hd[u]; i > 0; i = ed[i].nxt) {
edge &e = ed[i];
if (e.v != fa && !vis[e.v]) {
dfs4(e.v, D + e.w, u);
}
}
}
void dfs3(int u, LL D, int fa) {
dis[++cnt] = D;
for (int i = hd[u]; i > 0; i = ed[i].nxt) {
edge &e = ed[i];
if (e.v != fa && vis[e.v]) {
dfs4(e.v, 0, 0);
dfs3(e.v, D + e.w, u);
}
}
}
void clean() {
en = 0, ms(hd, -1), ms(vis, 0);
}
int solve() {
clean();
scanf("%d%d", &n, &s);
for (int x, y, w, i = 1; i < n; ++i) {
scanf("%d%d%d", &x, &y, &w);
ins(x, y, w), ins(y, x, w);
}
ms(dis, 0), dfs1(1, 0);
whw1 = 0;
for (int i = 1; i <= n; ++i) if (dis[whw1] < dis[i]) whw1 = i;
ms(dis, 0), dfs1(whw1, 0);
whw2 = 0;
for (int i = 1; i <= n; ++i) if (dis[whw2] < dis[i]) whw2 = i;
vis[whw1] = 1, dfs2(whw1, 0);
ms(dis, 0), cnt = 0, dfs3(whw1, 0, 0);
int l = 1;
LL ans = 4223372036854775808ll;
for (int r = 1; r <= cnt; ++r) {
while (l < r && dis[r] - dis[l] > s) ++l;
ans = min(ans, max(dis[l], max(dis[cnt] - dis[r], gg)));
}
cout << ans;
return 0;
}
}
int main() {
flyinthesky::solve();
return 0;
}
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