hdu 4123（树形dp+倍增）

Bob’s Race

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3741    Accepted Submission(s): 1206

Problem Description
Bob wants to hold a race to encourage people to do sports. He has got trouble in choosing the route. There are N houses and N - 1 roads in his village. Each road connects two houses, and all houses are connected together. To make the race more interesting, he requires that every participant must start from a different house and run AS FAR AS POSSIBLE without passing a road more than once. The distance difference between the one who runs the longest distance and the one who runs the shortest distance is called “race difference” by Bob. Bob does not want the “race difference”to be more than Q. The houses are numbered from 1 to N. Bob wants that the No. of all starting house must be consecutive. He is now asking you for help. He wants to know the maximum number of starting houses he can choose, by other words, the maximum number of people who can take part in his race.
Input
There are several test cases.
The first line of each test case contains two integers N and M. N is the number of houses, M is the number of queries.
The following N-1 lines, each contains three integers, x, y and z, indicating that there is a road of length z connecting house x and house y.
The following M lines are the queries. Each line contains an integer Q, asking that at most how many people can take part in Bob’s race according to the above mentioned rules and under the condition that the“race difference”is no more than Q.

The input ends with N = 0 and M = 0.

(N<=50000 M<=500 1<=x,y<=N 0<=z<=5000 Q<=10000000)
Output
For each test case, you should output the answer in a line for each query.
Sample Input
5 5
1 2 3
2 3 4
4 5 3
3 4 2
1
2
3
4
5
0 0
Sample Output
1
3
3
3
5
Source
2011 Asia Fuzhou Regional Contest
  
  1 #include<iostream>
  2 #include<cstdio>
  3 #include<algorithm>
  4 #include<cstring>
  5 #include<cstdlib>
  6 #include<string.h>
  7 #include<set>
  8 #include<vector>
  9 #include<queue>
 10 #include<stack>
 11 #include<map>
 12 #include<cmath>
 13 typedef long long ll;
 14 typedef unsigned long long LL;
 15 using namespace std;
 16 const double PI=acos(-1.0);
 17 const double eps=0.0000000001;
 18 const int N=50000+100;
 19 int head[N];
 20 int dp1[N][30],dp2[N][30];
 21 int tot;
 22 int n;
 23 struct node{
 24     int to,next,w;
 25 }edge[N<<1];
 26 int dp[N][3];
 27 int vis[N];
 28 int a[N];
 29 void init(){
 30     memset(head,-1,sizeof(head));
 31     memset(dp,0,sizeof(dp));
 32     tot=0;
 33 }
 34 void add(int u,int v,int w){
 35     edge[tot].to=v;
 36     edge[tot].next=head[u];
 37     edge[tot].w=w;
 38     head[u]=tot++;
 39 }
 40 void DFS(int u,int fa){
 41     int maxx1=0;
 42     int maxx2=0;
 43     for(int i=head[u];i!=-1;i=edge[i].next){
 44         int v=edge[i].to;
 45         int w=edge[i].w;
 46         if(v==fa)continue;
 47         DFS(v,u);
 48         int maxx=dp[v][0]+w;
 49         if(maxx>=maxx1){
 50             maxx2=maxx1;
 51             maxx1=maxx;
 52             vis[u]=v;
 53         }
 54         else if(maxx>maxx2){
 55             maxx2=maxx;
 56         }
 57        // cout<<u<<" "<<v<<" "<<maxx1<<" "<<maxx2<<endl;
 58     }
 59     dp[u][0]=maxx1;
 60     dp[u][1]=maxx2;
 61 }
 62 void DFS1(int u,int fa){
 63     for(int i=head[u];i!=-1;i=edge[i].next){
 64         int v=edge[i].to;
 65         int w=edge[i].w;
 66         if(v==fa)continue;
 67         if(vis[u]==v){
 68             dp[v][2]=max(dp[u][1]+w,dp[u][2]+w);
 69         }
 70         else{
 71             dp[v][2]=max(dp[u][0]+w,dp[u][2]+w);
 72         }
 73         DFS1(v,u);
 74     }
 75 }
 76 void init_RMQ(){
 77     for(int i=1;i<=n;i++){
 78         dp1[i][0]=a[i];
 79         dp2[i][0]=a[i];
 80     }
 81     for(int j=1;(1<<j)<=n;j++)
 82     for(int i=1;i+(1<<j)-1<=n;i++){
 83         dp1[i][j]=max(dp1[i][j-1],dp1[i+(1<<(j-1))][j-1]);
 84         dp2[i][j]=min(dp2[i][j-1],dp2[i+(1<<(j-1))][j-1]);
 85     }
 86 }
 87 int getRMQ(int L,int R){
 88     int k=0;
 89     while((1<<(k+1))<=R-L+1)k++;
 90     int maxx=max(dp1[L][k],dp1[R-(1<<k)+1][k]);
 91     int minn=min(dp2[L][k],dp2[R-(1<<k)+1][k]);
 92     return maxx-minn;
 93 }
 94 
 95 int main(){
 96     int m;
 97     while(scanf("%d%d",&n,&m)!=EOF){
 98 
 99         if(m==0&&n==0)break;
100         init();
101         for(int i=2;i<=n;i++){
102             int u,v,w;
103             scanf("%d%d%d",&u,&v,&w);
104             add(u,v,w);
105             add(v,u,w);
106         }
107         DFS(1,-1);
108         DFS1(1,-1);
109         for(int i=1;i<=n;i++){
110             a[i]=max(dp[i][0],dp[i][2]);
111             //cout<<a[i]<<" ";
112         }
113         init_RMQ();
114         for(int i=0;i<m;i++){
115             int x;
116             scanf("%d",&x);
117             int l=1;
118             int r=1;
119             int ans=1;
120             while(l<=n&&r<=n){
121                 while(r<=n&&getRMQ(l,r)<=x){
122                     r++;
123                 }
124                 ans=max(ans,r-l);
125                 l++;
126             }
127             cout<<ans<<endl;
128         }
129     }
130 }

 

posted on 2017-09-23 00:30  见字如面  阅读(114)  评论(0)  编辑  收藏  举报

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