hdu 4123(树形dp+倍增)
Bob’s Race
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3741 Accepted Submission(s): 1206
wants to hold a race to encourage people to do sports. He has got
trouble in choosing the route. There are N houses and N - 1 roads in his
village. Each road connects two houses, and all houses are connected
together. To make the race more interesting, he requires that every
participant must start from a different house and run AS FAR AS POSSIBLE
without passing a road more than once. The distance difference between
the one who runs the longest distance and the one who runs the shortest
distance is called “race difference” by Bob. Bob does not want the “race
difference”to be more than Q. The houses are numbered from 1 to N. Bob
wants that the No. of all starting house must be consecutive. He is now
asking you for help. He wants to know the maximum number of starting
houses he can choose, by other words, the maximum number of people who
can take part in his race.
The first line of each test case contains two integers N and M. N is the number of houses, M is the number of queries.
The
following N-1 lines, each contains three integers, x, y and z,
indicating that there is a road of length z connecting house x and house
y.
The following M lines are the queries. Each line contains an
integer Q, asking that at most how many people can take part in Bob’s
race according to the above mentioned rules and under the condition that
the“race difference”is no more than Q.
The input ends with N = 0 and M = 0.
(N<=50000 M<=500 1<=x,y<=N 0<=z<=5000 Q<=10000000)
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string.h>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<cmath>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const double PI=acos(-1.0);
const double eps=0.0000000001;
const int N=+;
int head[N];
int dp1[N][],dp2[N][];
int tot;
int n;
struct node{
int to,next,w;
}edge[N<<];
int dp[N][];
int vis[N];
int a[N];
void init(){
memset(head,-,sizeof(head));
memset(dp,,sizeof(dp));
tot=;
}
void add(int u,int v,int w){
edge[tot].to=v;
edge[tot].next=head[u];
edge[tot].w=w;
head[u]=tot++;
}
void DFS(int u,int fa){
int maxx1=;
int maxx2=;
for(int i=head[u];i!=-;i=edge[i].next){
int v=edge[i].to;
int w=edge[i].w;
if(v==fa)continue;
DFS(v,u);
int maxx=dp[v][]+w;
if(maxx>=maxx1){
maxx2=maxx1;
maxx1=maxx;
vis[u]=v;
}
else if(maxx>maxx2){
maxx2=maxx;
}
// cout<<u<<" "<<v<<" "<<maxx1<<" "<<maxx2<<endl;
}
dp[u][]=maxx1;
dp[u][]=maxx2;
}
void DFS1(int u,int fa){
for(int i=head[u];i!=-;i=edge[i].next){
int v=edge[i].to;
int w=edge[i].w;
if(v==fa)continue;
if(vis[u]==v){
dp[v][]=max(dp[u][]+w,dp[u][]+w);
}
else{
dp[v][]=max(dp[u][]+w,dp[u][]+w);
}
DFS1(v,u);
}
}
void init_RMQ(){
for(int i=;i<=n;i++){
dp1[i][]=a[i];
dp2[i][]=a[i];
}
for(int j=;(<<j)<=n;j++)
for(int i=;i+(<<j)-<=n;i++){
dp1[i][j]=max(dp1[i][j-],dp1[i+(<<(j-))][j-]);
dp2[i][j]=min(dp2[i][j-],dp2[i+(<<(j-))][j-]);
}
}
int getRMQ(int L,int R){
int k=;
while((<<(k+))<=R-L+)k++;
int maxx=max(dp1[L][k],dp1[R-(<<k)+][k]);
int minn=min(dp2[L][k],dp2[R-(<<k)+][k]);
return maxx-minn;
} int main(){
int m;
while(scanf("%d%d",&n,&m)!=EOF){ if(m==&&n==)break;
init();
for(int i=;i<=n;i++){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
DFS(,-);
DFS1(,-);
for(int i=;i<=n;i++){
a[i]=max(dp[i][],dp[i][]);
//cout<<a[i]<<" ";
}
init_RMQ();
for(int i=;i<m;i++){
int x;
scanf("%d",&x);
int l=;
int r=;
int ans=;
while(l<=n&&r<=n){
while(r<=n&&getRMQ(l,r)<=x){
r++;
}
ans=max(ans,r-l);
l++;
}
cout<<ans<<endl;
}
}
}
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