Bob’s Race

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1994    Accepted Submission(s): 619

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  

首先是两遍dfs,预处理出每个结点到叶子结点的巨大距离。

然后使用rmq来查询区间的最大最小值。

每次查询扫描一遍就可以了、

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013-11-8 16:56:11
  4 File Name     :E:\2013ACM\专题强化训练\区域赛\2011福州\C.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 const int MAXN = 50010;
 21 struct Edge
 22 {
 23     int to,next;
 24     int w;
 25 }edge[MAXN*2];
 26 int head[MAXN],tot;
 27 void init()
 28 {
 29     tot = 0;
 30     memset(head,-1,sizeof(head));
 31 }
 32 void addedge(int u,int v,int w)
 33 {
 34     edge[tot].to = v;
 35     edge[tot].w = w;
 36     edge[tot].next = head[u];
 37     head[u] = tot++;
 38 }
 39 int maxn[MAXN],smaxn[MAXN];
 40 int maxid[MAXN],smaxid[MAXN];
 41 void dfs1(int u,int pre)
 42 {
 43     maxn[u] = smaxn[u] = maxid[u] = smaxid[u] = 0;
 44     for(int i = head[u];i != -1;i = edge[i].next)
 45     {
 46         int v = edge[i].to;
 47         if(pre == v)continue;
 48         dfs1(v,u);
 49         if(maxn[v] + edge[i].w > smaxn[u])
 50         {
 51             smaxid[u] = v;
 52             smaxn[u] = maxn[v] + edge[i].w;
 53             if(maxn[u] < smaxn[u])
 54             {
 55                 swap(maxn[u],smaxn[u]);
 56                 swap(maxid[u],smaxid[u]);
 57             }
 58         }
 59     }
 60 }
 61 void dfs2(int u,int pre)
 62 {
 63     for(int i = head[u];i != -1;i = edge[i].next)
 64     {
 65         int v = edge[i].to;
 66         if(pre == v)continue;
 67         if(maxid[u] == v)
 68         {
 69             if(smaxn[u] + edge[i].w > smaxn[v])
 70             {
 71                 smaxn[v] = smaxn[u] + edge[i].w;
 72                 smaxid[v] = u;
 73                 if(maxn[v] < smaxn[v])
 74                 {
 75                     swap(maxn[v],smaxn[v]);
 76                     swap(maxid[v],smaxid[v]);
 77                 }
 78             }
 79         }
 80         else
 81         {
 82             if(maxn[u] + edge[i].w > smaxn[v])
 83             {
 84                 smaxn[v] = maxn[u] + edge[i].w;
 85                 smaxid[v] = u;
 86                 if(maxn[v] < smaxn[v])
 87                 {
 88                     swap(maxn[v],smaxn[v]);
 89                     swap(maxid[v],smaxid[v]);
 90                 }
 91             }
 92         }
 93         dfs2(v,u);
 94     }
 95 }
 96 int a[MAXN];
 97 
 98 int dp1[MAXN][20];
 99 int dp2[MAXN][20];
100 int mm[MAXN];
101 void initRMQ(int n)
102 {
103     mm[0] = -1;
104     for(int i = 1;i <= n;i++)
105     {
106         mm[i] = ((i&(i-1)) == 0)?mm[i-1]+1:mm[i-1];
107         dp1[i][0] = a[i];
108         dp2[i][0] = a[i];
109     }
110     for(int j = 1;j <= mm[n];j++)
111         for(int i = 1;i + (1<<j) - 1 <= n;i++)
112         {
113             dp1[i][j] = max(dp1[i][j-1],dp1[i + (1<<(j-1))][j-1]);
114             dp2[i][j] = min(dp2[i][j-1],dp2[i + (1<<(j-1))][j-1]);
115         }
116 }
117 int rmq(int x,int y)
118 {
119     int k = mm[y-x+1];
120     return max(dp1[x][k],dp1[y-(1<<k)+1][k]) - min(dp2[x][k],dp2[y-(1<<k)+1][k]);
121 }
122 int main()
123 {
124     //freopen("in.txt","r",stdin);
125     //freopen("out.txt","w",stdout);
126     int n,m;
127     int u,v,w;
128     int Q;
129     while(scanf("%d%d",&n,&m) == 2)
130     {
131         if(n == 0 && m == 0)break;
132         init();
133         for(int i = 1;i < n;i++)
134         {
135             scanf("%d%d%d",&u,&v,&w);
136             addedge(u,v,w);
137             addedge(v,u,w);
138         }
139         dfs1(1,1);
140         dfs2(1,1);
141         for(int i = 1;i <= n;i++)
142             a[i] = maxn[i];
143         initRMQ(n);
144         while(m--)
145         {
146             scanf("%d",&Q);
147             int ans = 0;
148             int id = 1;
149             for(int i = 1;i <= n;i++)
150             {
151                 while(id <= i && rmq(id,i) > Q)id++;
152                 ans = max(ans,i-id+1);
153             }
154             printf("%d\n",ans);
155         }
156     }
157     return 0;
158 }