POJ1741 Tree + BZOJ1468 Tree 【点分治】
POJ1741 Tree + BZOJ1468 Tree
Description
Give a tree with n vertices,each edge has a length(positive integer less than 1001).
Define dist(u,v)=The min distance between node u and v.
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k.
Write a program that will count how many pairs which are valid for a given tree.
Input
The input contains several test cases. The first line of each test case contains two integers n, k. (n<=10000) The following n-1 lines each contains three integers u,v,l, which means there is an edge between node u and v of length l.
The last test case is followed by two zeros.
Output
For each test case output the answer on a single line.
Sample Input
5 4
1 2 3
1 3 1
1 4 2
3 5 1
0 0
Sample Output
8
BZOJ上的和POJ上的其实差不多，只是输入输出格式的区别
然后就是点分治的板子
但是dcy说这题卡常，我也不知道为什么
我也没有卡啊
POJ1741：
#include<iostream>
#include<vector>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define N 40010
int n,k,cnt,rt,siz_tree,ans;
int siz[N],F[N],dis[N],head[N],val[N];
bool vis[N];
struct Edge{int v,w,next;}E[N<<1];
void init(){
cnt=0;ans=0;
memset(head,0,sizeof(head));
memset(dis,0,sizeof(dis));
memset(vis,0,sizeof(vis));//******
}
void add(int u,int v,int w){
cnt++;
E[cnt].v=v;
E[cnt].w=w;
E[cnt].next=head[u];
head[u]=cnt;
}
void getroot(int u,int fa){
siz[u]=1;F[u]=0;
for(int i=head[u];i;i=E[i].next){
int v=E[i].v;
if(vis[v]||v==fa)continue;
getroot(v,u);
siz[u]+=siz[v];
F[u]=max(F[u],siz[v]);
}
F[u]=max(F[u],siz_tree-siz[u]);
if(F[u]<F[rt])rt=u;
}
void getdis(int u,int fa){
val[++val[0]]=dis[u];
for(int i=head[u];i;i=E[i].next){
int v=E[i].v;
if(v==fa||vis[v])continue;
dis[v]=dis[u]+E[i].w;
getdis(v,u);
}
}
int work(int u,int pre){
val[0]=0;dis[u]=pre;
getdis(u,0);
sort(val+1,val+val[0]+1);
int l=1,r=val[0],res=0;
while(l<r){
if(val[r]+val[l]<=k)res+=r-l,l++;
else r--;
}
return res;
}
void solve(int u){
ans+=work(u,0);
vis[u]=1;
for(int i=head[u];i;i=E[i].next){
int v=E[i].v;
if(vis[v])continue;
ans-=work(v,E[i].w);
F[rt=0]=siz_tree=siz[v];
getroot(v,0);
solve(rt);
}
}
int main(){
while(1){
scanf("%d%d",&n,&k);
if(!n&&!k)return 0;
init();
for(int i=1;i<n;i++){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
F[rt=0]=siz_tree=n;
getroot(1,0);
solve(rt);
printf("%d\n",ans);
}
return 0;
}BZOJ1468：
#include<iostream>
#include<vector>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define N 40010
int n,k,cnt,rt,siz_tree,ans;
int siz[N],F[N],dis[N],head[N],val[N];
bool vis[N];
struct Edge{int v,w,next;}E[N<<1];
void init(){
cnt=0;ans=0;
memset(head,0,sizeof(head));
memset(dis,0,sizeof(dis));
memset(vis,0,sizeof(vis));//******
}
void add(int u,int v,int w){
cnt++;
E[cnt].v=v;
E[cnt].w=w;
E[cnt].next=head[u];
head[u]=cnt;
}
void getroot(int u,int fa){
siz[u]=1;F[u]=0;
for(int i=head[u];i;i=E[i].next){
int v=E[i].v;
if(vis[v]||v==fa)continue;
getroot(v,u);
siz[u]+=siz[v];
F[u]=max(F[u],siz[v]);
}
F[u]=max(F[u],siz_tree-siz[u]);
if(F[u]<F[rt])rt=u;
}
void getdis(int u,int fa){
val[++val[0]]=dis[u];
for(int i=head[u];i;i=E[i].next){
int v=E[i].v;
if(v==fa||vis[v])continue;
dis[v]=dis[u]+E[i].w;
getdis(v,u);
}
}
int work(int u,int pre){
val[0]=0;dis[u]=pre;
getdis(u,0);
sort(val+1,val+val[0]+1);
int l=1,r=val[0],res=0;
while(l<r){
if(val[r]+val[l]<=k)res+=r-l,l++;
else r--;
}
return res;
}
void solve(int u){
ans+=work(u,0);
vis[u]=1;
for(int i=head[u];i;i=E[i].next){
int v=E[i].v;
if(vis[v])continue;
ans-=work(v,E[i].w);
F[rt=0]=siz_tree=siz[v];
getroot(v,0);
solve(rt);
}
}
int main(){
scanf("%d",&n);
init();
for(int i=1;i<n;i++){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
scanf("%d",&k);
F[rt=0]=siz_tree=n;
getroot(1,0);
solve(rt);
printf("%d\n",ans);
return 0;
}
