Mart Master II
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 285 Accepted Submission(s): 94
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?
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.
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
题目解说：http://www.itnose.net/detail/6118094.html
代码：
题目解说：http://www.itnose.net/detail/6118094.html
代码：
/* ***********************************************
Author :rabbit
Created Time :2014/11/1 22:13:28
File Name :6.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=100100;
struct Edge{
int next,to,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].next=head[u];
edge[tot].w=w;
head[u]=tot++;
}
int size[maxn],vis[maxn],fa[maxn],que[maxn];
int TT;
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],np[maxn];
int dis[maxn],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("data.in","r",stdin);
//freopen("data.out","w",stdout);
int n;
memset(vis,0,sizeof(vis));
TT=0;
while(~scanf("%d",&n)){
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]);
queue<int> q;
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()){
int u=q.front();
q.pop();
ff[u]=0;
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]=1;
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]);
cout<<ret<<endl;
}
return 0;
}