HDU5016 树分治。题意就是要找一个点，使得可以占领的点最多。

Mart Master II

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 156 Accepted Submission(s): 58

Problem Description

Trader Dogy lives in city S, which consists of n districts. There are n - 1 bidirectional roads in city S, each connects a pair of districts. Indeed, city S is connected, i.e. people can travel between every pair of districts by roads. In some districts there are marts founded by Dogy’s competitors. when people go to marts, they’ll choose the nearest one. In cases there are more than one nearest marts, they’ll choose the one with minimal city number. Dogy’s money could support him to build only one new marts, he wants to attract as many people as possible, that is, to build his marts in some way that maximize the number of people who will choose his mart as favorite. Could you help him?

Input

There are multiple test cases. Please process till EOF. In each test case: First line: an integer n indicating the number of districts. Next n - 1 lines: each contains three numbers bi, ei and wi, (1 ≤ bi,ei ≤ n,1 ≤ wi ≤ 10000), indicates that there’s one road connecting city bi and ei, and its length is wi. Last line : n(1 ≤ n ≤ 105) numbers, each number is either 0 or 1, i-th number is 1 indicates that the i-th district has mart in the beginning and vice versa.

Output

For each test case, output one number, denotes the number of people you can attract, taking district as a unit.

Sample Input

5 1 2 1 2 3 1 3 4 1 4 5 1 1 0 0 0 1 5 1 2 1 2 3 1 3 4 1 4 5 1 1 0 0 0 0 1 1 1 0

Sample Output

2 4 0 1

Source

2014 ACM/ICPC Asia Regional Xi’an Online

找到树根。在一颗子树的，进行分治处理。然后不在一颗子树的，要经过树根。就是 dis[A] + dis[B] < WW[B]. 这样A就是可以统治B的。

/* ***
Author :kuangbin
Created Time :2014/9/23 11:40:11
File Name :E:\2014ACM\专题学习\树分治\HDU5016.cpp
************************************************ */

#include <stdio.h>

#include <string.h>

#include

#include <math.h>

#include <stdlib.h>

#include <time.h>
using namespace std;
const int MAXN = 100010;
const int INF = 0x3f3f3f3f;
struct Edge{
int to,next,w;
}edge[MAXN*2];
int head[MAXN],tot;
void init(){
tot = 0;
memset(head,-1,sizeof(head));
}
inline void addedge(int u,int v,int w){
edge[tot].to = v;
edge[tot].w = w;
edge[tot].next = head[u];
head[u] = tot++;
}
int size[MAXN],vis[MAXN],fa[MAXN],que[MAXN];
int TT;//时间戳
//找重心
inline int getroot(int u){
int Min = MAXN, root = 0;
int l,r;
que[l = r = 1] = u;
fa[u] = 0;
for(;l <= r;l++)
for(int i = head[que[l]]; i != -1;i = edge[i].next){
int v = edge[i].to;
if(v == fa[que[l]] || vis[v] == TT)continue;
que[++r] = v;
fa[v] = que[l];
}
for(l–;l;l–){
int x = que[l], Max = 0;
size[x] = 1;
for(int i = head[x];i != -1;i = edge[i].next){
int v = edge[i].to;
if(v == fa[x] || vis[v] == TT)continue;
Max = max(Max,size[v]);
size[x] += size[v];
}
Max = max(Max,r - size[x]);
if(Max < Min){
Min = Max; root = x;
}
}
return root;
}

int ans[MAXN];
pair<int,int>pp[MAXN];
pair<int,int>np[MAXN];
int dis[MAXN];
int type[MAXN];
inline void go(int u,int pre,int w,int tt){
int l,r;
que[l = r = 1] = u;
fa[u] = pre; dis[u] = w;
for(;l <= r;l++)
for(int i = head[que[l]];i != -1;i = edge[i].next){
int v = edge[i].to;
if(v == fa[que[l]] || vis[v] == TT)continue;
que[++r] = v;
fa[v] = que[l];
dis[v] = dis[que[l]] + edge[i].w;
}
int cnt = 0;
for(int i = 1;i <= r;i++){
int x = que[i];
pp[cnt++] = make_pair(np[x].first-dis[x],np[x].second);
}
sort(pp,pp+cnt);
for(int i = 1;i <= r;i++){
int x = que[i];
if(type[x])continue;
int id = lower_bound(pp,pp+cnt,make_pair(dis[x],x)) - pp;
ans[x] += (cnt-id)*tt;
}
}
void solve(int u){
int root = getroot(u);
vis[root] = TT;
go(root,0,0,1);
for(int i = head[root];i != -1;i = edge[i].next){
int v = edge[i].to;
if(vis[v] == TT)continue;
go(v,root,edge[i].w,-1);
}
for(int i = head[root];i != -1;i = edge[i].next){
int v = edge[i].to;
if(vis[v] == TT)continue;
solve(v);
}
}
bool ff[MAXN];
int main()
{
//freopen(“in.txt”,”r”,stdin);
//freopen(“out.txt”,”w”,stdout);
int n;
memset(vis,0,sizeof(vis));
TT = 0;
while(scanf(“%d”,&n) == 1){
init();
int u,v,w;
for(int i = 1;i < n;i++){
scanf(“%d%d%d”,&u,&v,&w);
addedge(u,v,w);
addedge(v,u,w);
}
for(int i = 1;i <= n;i++)scanf(“%d”,&type[i]);
queueq;
for(int i = 1;i <= n;i++){
if(type[i]){
np[i] = make_pair(0,i);
ff[i] = true;
q.push(i);
}
else{
np[i] = make_pair(INF,0);
ff[i] = false;
}
}
while(!q.empty()){
u = q.front();
q.pop();
ff[u] = false;
for(int i = head[u];i != -1;i = edge[i].next){
v = edge[i].to;
pair<int,int>tmp = make_pair(np[u].first+edge[i].w,np[u].second);
if(tmp < np[v]){
np[v] = tmp;
if(!ff[v]){
ff[v] = true;
q.push(v);
}
}
}
}
TT++;
for(int i = 1;i <= n;i++)ans[i] = 0;
solve(1);
int ret = 0;
for(int i = 1;i <= n;i++)ret = max(ret,ans[i]);
printf(“%d\n”,ret);
}
return 0;
}