HDU 4916 tree governance, hdu4916 tree governance

HDU 4916 tree governance, hdu4916 tree governance
Mart Master IITime Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission (s): 285 Accepted Submission (s): 94


Problem DescriptionTrader 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 cocould support him to build onlyOneNew marts, he wants to attract as your 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. cocould you help him?
InputThere 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.
OutputFor each test case, output one number, denotes the number of people you can attract, taking district as a unit.
Sample Input

51 2 12 3 13 4 14 5 11 0 0 0 151 2 12 3 13 4 14 5 11 0 0 0 01110

 
Sample Output

2401

 
Source2014 ACM/ICPC Asia Regional Xi 'an Online
Http://www.itnose.net/detail/6118094.html
Code:

/* ***********************************************Author :rabbitCreated Time :2014/11/1 22:13:28File 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;}

HDU 1007 Quoit Design governance solution recently provides guidance on various timeout Models

Try changing the function to inline ?.... I sorted y by base. Me too, tle...