Distance Statistics
Description
Frustrated at the number of distance queries required to find a reasonable route for his cow marathon, FJ decides to ask queries from which he can learn more information. Specifically, he supplies an integer K (1 <= K <= 1,000,000,000) and wants to know how many pairs of farms lie at a distance at most K from each other (distance is measured in terms of the length of road required to travel from one farm to another). Please only count pairs of distinct farms (i.e. do not count pairs such as (farm #5, farm #5) in your answer).
Input
* Lines 1 ..M+1: Same input format as in "Navigation Nightmare"
* Line M+2: A single integer, K.
Output
* Line 1: The number of pairs of farms that are at a distance of at most K from each-other.
Sample Input
7 6 1 6 13 E 6 3 9 E 3 5 7 S 4 1 3 N 2 4 20 W 4 7 2 S 10
Sample Output
5
Hint
There are 5 roads with length smaller or equal than 10, namely 1-4 (3), 4-7 (2), 1-7 (5), 3-5 (7) and 3-6 (9).
题解：
POJ 1741
http://www.cnblogs.com/zxhl/p/5692688.html
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int N = 4e4+20, M = 1e2+10, mod = 1e9+7, inf = 1e9+1000;
typedef long long ll;
int ans, n,m,root , t = 1,K,siz[N],head[N],f[N],deep[N],d[N],allnode,vis[N];
struct edg{int to,next,v,w;}e[N * 4];
void add(int u,int v,int w) {e[t].to=v;e[t].v=w;e[t].next=head[u];head[u]=t++;}
void getroot(int x,int fa) {
siz[x] = 1;
f[x] = 0;
for(int i=head[x];i;i=e[i].next) {
int to = e[i].to;
if(to == fa || vis[to]) continue;
getroot(to,x);
siz[x] += siz[to];
f[x] = max(f[x] , siz[to]);
}
f[x] = max(f[x] , allnode - siz[x]);
if(f[x] < f[root]) root = x;
}
void getdeep(int x,int fa) {
if(d[x] <= K) deep[++deep[0]]=d[x];
for(int i=head[x];i;i=e[i].next) {
int to = e[i].to;
if(to == fa || vis[to]) continue;
d[to] = d[x] + e[i].v;
getdeep(to,x);
}
}
int cal(int x,int now) {
d[x]=now;deep[0] = 0;
getdeep(x,0);
sort(deep+1,deep+deep[0]+1);
int all = 0;
for(int l=1,r=deep[0];l<r;) {
if(deep[l]+deep[r] <= K) {all+=r-l;l++;}
else r--;
}
return all;
}
void work(int x) {
ans+=cal(x,0);
vis[x] = 1;
for(int i=head[x];i;i=e[i].next) {
int to = e[i].to;
if(vis[to]) continue;
ans-=cal(to,e[i].v);
allnode = siz[to];
root = 0;
getroot(to,root);
work(root);
}
}
void init()
{
memset(head,0,sizeof(head));
t = 1;
ans = root = 0;
memset(vis,0,sizeof(vis));
}
int main()
{
while(~scanf("%d%d",&n,&m)) {
init();
for(int i=1;i<n;i++) {
int a,b,c;char ch[2];
scanf("%d%d%d%s",&a,&b,&c,ch);
add(a,b,c) , add(b,a,c);
}
scanf("%d",&K);
allnode=n;f[0]=inf;
getroot(1,0);
work(root);
printf("%d\n",ans);
}
}