poj 2114 Boatherds (树分治)

链接:http://poj.org/problem?id=2114

题意:

求树上距离为k的点对数量;

思路:

点分治。。

实现代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 0x7fffffff
const int M = 1e5+; struct node{
int to,next,w;
}e[M<<]; int n,m;
int vis[M],dis[M],d[M],siz[M],f[M],cnt,head[M],sum,root,k,ans; void init(){
cnt = ;
ans = ;
memset(head,,sizeof(head));
memset(vis,,sizeof(vis));
} void add(int u,int v,int w){
e[++cnt].to = v;e[cnt].w = w;e[cnt].next = head[u];head[u] = cnt;
} void get_root(int u,int fa){
siz[u] = ; f[u] = ;
for(int i = head[u];i;i=e[i].next){
int v = e[i].to;
if(v != fa&&!vis[v]){
get_root(v,u);
siz[u] += siz[v];
f[u] = max(f[u],siz[u]);
}
}
f[u] = max(f[u],sum - siz[u]);
if(f[u] < f[root]) root = u;
return ;
} void get_dis(int u,int fa){
if(d[u] <= k) dis[++dis[]] = d[u];
for(int i = head[u];i;i=e[i].next){
int v = e[i].to;
if(v != fa&&!vis[v]){
d[v] = d[u] + e[i].w;
get_dis(v,u);
}
}
return ;
} int cal(int u,int c){
d[u] = c; dis[] = ;
get_dis(u,);
sort(dis+,dis+dis[]+);
int l = ,r = dis[],ans = ;
while(l < r){
if(dis[l] + dis[r] < k) l++;
else if(dis[l] + dis[r] > k) r--;
else{
if(dis[l] == dis[r]){
ans += (r-l+)*(r-l)/;
break;
}
else {
int i = l,j = r;
while(dis[i] == dis[l]) i++;
while(dis[j] == dis[r]) j--;
ans += (i-l)*(r-j);
l = i; r = j;
}
}
}
return ans;
} void solve(int u){
ans += cal(u,);
vis[u] = ;
for(int i = head[u];i;i=e[i].next){
int v = e[i].to;
if(vis[v]) continue;
ans -= cal(v,e[i].w);
sum = siz[v];
root = ;
get_root(v,);
solve(root);
}
} int main()
{
int u,v,w;
while(scanf("%d",&n)&&n){
init();
for(int i = ;i <= n;i ++){
int x,v;
while(scanf("%d",&x)&&x){
scanf("%d",&v);
add(i,x,v); add(x,i,v);
}
}
int x;
while(scanf("%d",&x)&&x){
memset(vis,,sizeof(vis));
k = x; ans = root = ;
sum = n; f[] = inf;
get_root(,);
solve(root);
if(ans) printf("AYE\n");
else printf("NAY\n");
}
printf(".\n");
}
}
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