HDU 4123 Bob’s Race 树的直径+单调队列

题意:

给定n个点的带边权树Q个询问。

以下n-1行给出树

以下Q行每行一个数字表示询问。

首先求出dp[N] :dp[i]表示i点距离树上最远点的距离

询问u, 表示求出 dp 数组中最长的连续序列使得序列中最大值-最小值 <= u,输出这个序列的长度。

思路:

求dp数组就是求个树的直径然后dfs一下。

对于每一个询问,能够用一个单调队列维护一下。O(n)的回答。

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <queue>
#include <algorithm>
#include <cmath>
using namespace std;
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if(c=getchar(),c==EOF) return 0;
while(c!='-'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
ret*=sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x <0) {
putchar('-');
x = -x;
}
if(x>9) pt(x/10);
putchar(x%10+'0');
}
typedef long long ll;
const int N = 50010;
int n, Q;
struct Edge{
int to, nex; ll dis;
}edge[N<<1];
struct node {
int v, id;
node() {}
node(int _id, int _v) {
id = _id; v = _v;
}
};
int head[N], edgenum;
void init(){for(int i = 1; i <= n; i++)head[i] = -1; edgenum = 0;}
void add(int u, int v, ll d){
Edge E = {v, head[u], d};
edge[edgenum] = E;
head[u] = edgenum++;
}
ll dis[N], dp[N], len;
int Stack[N], top, pre[N], vis[N];
int BFS(int x){
for(int i = 1; i <= n; i++)
dis[i] = -1;
dis[x] = 0; pre[x] = -1;
int far = x;
queue<int> q; q.push(x);
while(!q.empty())
{
int u = q.front(); q.pop();
for(int i = head[u]; ~i; i = edge[i].nex){
int v = edge[i].to;
if(dis[v] == -1)
{
dis[v] = dis[u] + edge[i].dis;
pre[v] = u;
if(dis[far] < dis[v])
far = v;
q.push(v);
}
}
}
return far;
}
void dfs(int u){
vis[u] = 1;
for(int i = head[u]; ~i; i = edge[i].nex)
{
int v = edge[i].to;
if(vis[v])continue;
dp[v] = dp[u] + edge[i].dis;
dfs(v);
}
}
void build(){//预处理树的直径
int E = BFS(1);
int S = BFS(E);
top = 0;
int u = S;
len = dis[S];
for(int i = 1; i <= n; i++) vis[i] = 0;
while(u!=-1)
{
Stack[top++] = u;
dp[u] = max(dis[u], len - dis[u]);
vis[u] = 1;
u = pre[u];
}
for(int i = 0; i < top; i++) dfs(Stack[i]);
}
void input(){
init(); ll d;
for(int i = 1, u, v; i < n; i++)
{
rd(u); rd(v); rd(d);
add(u, v, d); add(v, u, d);
}
} node mx[N], mi[N];
int h1, t1, h2, t2;
int main() {
int v, idx, ans;
while(cin>>n>>Q, n+Q) {
input();
build();
while(Q--)
{
rd(v);
ans = h1 = t1 = h2 = t2 = 0;
idx = 1;
for (int i = 1; i <= n; ++i) {
while (h1!=t1 && mx[t1-1].v <= dp[i])
-- t1;
mx[t1++] = node(i, dp[i]);
while (h2!=t2 && mi[t2-1].v >= dp[i])
-- t2;
mi[t2++] = node(i, dp[i]);
while (h1!=t1&&h2!=t2) {
if (mx[h1].v-mi[h2].v>v)
++ idx;
else
break;
while (h1!=t1&&mx[h1].id<idx)
++h1;
while (h2!=t2&&mi[h2].id<idx)
++h2;
}
ans = max(ans, i-idx+1);
}
pt(ans);
putchar('\n');
}
}
return 0;
}
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