题意:给一棵树求之间距离不超过k的点对数量。。
和poj1741一模一样。。。一样的做法。。每次找经过根的点对的数量。。统计数量时,排个序然后头尾开始找数量就好
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=40010;
const int INF=1<<30;
struct EDGE
{
int v,next;
int dist;
}edge[MAXN<<1];
int head[MAXN],size;
void init()
{
memset(head,-1,sizeof(head));
size=0;
}
void add_edge(int u,int v,int c)
{
edge[size].v=v;
edge[size].dist=c;
edge[size].next=head[u];
head[u]=size++;
}
int root,siz[MAXN],num[MAXN],tot_size;
bool vis[MAXN];
void get_root(int u,int fa)
{
num[u]=0;
siz[u]=1;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(v==fa||vis[v])
continue;
get_root(v,u);
siz[u]+=siz[v];
num[u]=max(num[u],siz[v]);
}
num[u]=max(num[u],tot_size-num[u]);
if(num[root]>num[u])
root=u;
}
int a[MAXN],cnt,k,dep[MAXN];
void get_dep(int u,int fa)
{
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(v==fa||vis[v])
continue;
dep[v]=dep[u]+edge[i].dist;
a[cnt++]=dep[v];
get_dep(v,u);
}
}
int get_num(int u,int val)
{
cnt=0;
dep[u]=val;
a[cnt++]=val;
get_dep(u,-1);
sort(a,a+cnt);
int l=0,r=cnt-1;
int res=0;
while(l<=r)
{
if(a[l]+a[r]<=k)
{
res+=r-l;
l++;
}
else
r--;
}
return res;
}
int ans;
void dfs(int u)
{
vis[u]=1;
ans+=get_num(u,0);
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(vis[v])
continue;
ans-=get_num(v,edge[i].dist);
}
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(vis[v])
continue;
root=0;
get_root(v,-1);
tot_size=siz[v];
dfs(root);
}
}
int main()
{
int n,m,i;
while(scanf("%d%d",&n,&m)==2)
{
int u,v,c;
char op[2];
init();
while(m--)
{
scanf("%d%d%d%s",&u,&v,&c,op);
add_edge(u,v,c);
add_edge(v,u,c);
}
scanf("%d",&k);
memset(num,0,sizeof(num));
root=0;
ans=0;
num[0]=INF;
tot_size=n;
get_root(1,-1);
dfs(root);
printf("%d\n",ans);
}
return 0;
}