标签：ring 技术分享 algorithm which training sed pre art instr
| Time Limit: 10000MS | Memory Limit: 131072K | |
| Total Submissions: 4123 | Accepted: 1029 | |
| Case Time Limit: 3000MS | ||
Description
flymouse’s sister wc is very capable at sports and her favorite event is walking race. Chasing after the championship in an important competition, she comes to a training center to attend a training course. The center has N check-points numbered 1 through N. Some pairs of check-points are directly connected by two-way paths. The check-points and the paths form exactly a tree-like structure. The course lasts N days. On the i-th day, wc picks check-point i as the starting point and chooses another check-point as the finishing point and walks along the only simple path between the two points for the day’s training. Her choice of finishing point will make it that the resulting path will be the longest among those of all possible choices.
After every day’s training, flymouse will do a physical examination from which data will obtained and analyzed to help wc’s future training be better instructed. In order to make the results reliable, flymouse is not using data all from N days for analysis. flymouse’s model for analysis requires data from a series of consecutive days during which the difference between the longest and the shortest distances wc walks cannot exceed a bound M. The longer the series is, the more accurate the results are. flymouse wants to know the number of days in such a longest series. Can you do the job for him?
Input
The input contains a single test case. The test case starts with a line containing the integers N (N ≤ 106) and M (M < 109). Then follow N − 1 lines, each containing two integers fi and di (i = 1, 2, …, N − 1), meaning the check-points i + 1 and fi are connected by a path of length di.
Output
Output one line with only the desired number of days in the longest series.
Sample Input
3 2 1 1 1 3
Sample Output
3
Hint
Explanation for the sample:
There are three check-points. Two paths of lengths 1 and 3 connect check-points 2 and 3 to check-point 1. The three paths along with wc walks are 1-3, 2-1-3 and 3-1-2. And their lengths are 3, 4 and 4. Therefore data from all three days can be used for analysis.
Source
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> #include<queue> #include<stack> #include<map> #include<stack> #include<set> #include<bitset> using namespace std; #define PI acos(-1.0) #define eps 1e-8 typedef long long ll; typedef pair<int,int > P; const int N=1e6+100,M=2e6+100; const int inf=0x3f3f3f3f; const ll INF=1e13+7,mod=1e9+7; struct edge { int from,to; ll w; int next; }; edge es[M]; int cnt,head[N]; ll dp[N][5]; void init() { cnt=0; memset(head,-1,sizeof(head)); } void addedge(int u,int v,ll w) { cnt++; es[cnt].from=u,es[cnt].to=v; es[cnt].w=w; es[cnt].next=head[u]; head[u]=cnt; } void dfs1(int u,int fa) { dp[u][0]=dp[u][1]=0; for(int i=head[u]; i!=-1; i=es[i].next) { edge e=es[i]; if(e.to==fa) continue; dfs1(e.to,u); ll tmp=dp[e.to][0]+e.w; if(tmp>dp[u][0]) swap(tmp,dp[u][0]); if(tmp>dp[u][1]) swap(tmp,dp[u][1]); } } void dfs2(int u,int fa) { for(int i=head[u]; i!=-1; i=es[i].next) { edge e=es[i]; if(e.to==fa) continue; if(dp[e.to][0]+e.w==dp[u][0]) dp[e.to][2]=max(dp[u][2],dp[u][1])+e.w; else dp[e.to][2]=max(dp[u][2],dp[u][0])+e.w; dfs2(e.to,u); } } ll Tree1[N<<2],Tree2[N<<2]; void pushup(int pos) { Tree1[pos]=max(Tree1[pos<<1],Tree1[pos<<1|1]); Tree2[pos]=min(Tree2[pos<<1],Tree2[pos<<1|1]); } void build(int l,int r,int pos) { if(l==r) { Tree1[pos]=Tree2[pos]=max(dp[l][0],dp[l][2]); return; } int mid=(l+r)>>1; build(l,mid,pos<<1); build(mid+1,r,pos<<1|1); pushup(pos); } void query(ll &ans1,ll &ans2,int L,int R,int l,int r,int pos) { if(L<=l&&r<=R) { ans1=max(ans1,Tree1[pos]); ans2=min(ans2,Tree2[pos]); return; } int mid=(l+r)>>1; if(L<=mid) query(ans1,ans2,L,R,l,mid,pos<<1); if(R>mid) query(ans1,ans2,L,R,mid+1,r,pos<<1|1); } int main() { int n; ll m; scanf("%d%lld",&n,&m); init(); for(int i=2,j; i<=n; i++) { ll w; scanf("%d%lld",&j,&w); addedge(i,j,w),addedge(j,i,w); } memset(dp,0,sizeof(dp)); dfs1(1,0); dfs2(1,0); int ans=0; build(1,n,1); for(int i=1,j=1; i<=n&&j<=n;) { ll maxx=-INF,minx=INF; query(maxx,minx,i,j,1,n,1); if(maxx-minx<=m) ans=max(ans,j-i+1),j++; else i++; if(n-i<ans) break; } printf("%d\n",ans); return 0; }
标签：ring 技术分享 algorithm which training sed pre art instr
原文地址：http://www.cnblogs.com/GeekZRF/p/7620211.html
