【题解】poj 3162 Walking Race 树形dp

题目描述

Walking Race
Time Limit: 10000MS Memory Limit: 131072K
Total Submissions: 4941 Accepted: 1252
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 ≤ 10^{6}) and M (M < \10^{9}). 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

题目大意

题目大概是给一棵n个结点边带权的树，记结点i到其他结点最远距离为d[i]，问d数组构成的这个序列中满足其中最大值与最小值的差不超过m的连续子序列最长是多长。

思路

各个结点到其他结点的最远距离可以用树形DP解决，HDU2196

因为有最大值和最小值，需要两个单调队列，一个维护最大值qmax，另一个维护最小值qmin

具体操作看代码

代码

#include<cmath>
#include<deque>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#define re register int
using namespace std;
inline int read(){
	int x=0,w=1;
	char ch=getchar();
	while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
	if(ch=='-') w=-1,ch=getchar();
	while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-48,ch=getchar();
	return x*w;
}
const int MAXN=1e6+200;
struct edge {
    int to,next,w;
} edges[MAXN*2];
int tot,dis[MAXN];
int head[MAXN];
void add_edge(int u,int v,int w) {
    edges[tot].to=v;
    edges[tot].w=w;
    edges[tot].next = head[u];
    head[u] = tot++;
}
int dist[MAXN][3];  //dist[i][0,1,2]分别为正向最大距离，正向次大距离，反向最大距离
int longest[MAXN];
int dfs1(int u,int fa) {
    if(dist[u][0]>=0) return dist[u][0];
    dist[u][0]=dist[u][1]=dist[u][2]=longest[u]=0;
    for(re i=head[u];i!=-1;i=edges[i].next) {
        int v=edges[i].to;
        if(v==fa) continue;
        if(dist[u][0]<dfs1(v,u)+edges[i].w) {
            longest[u]=v;
            dist[u][1]=max(dist[u][1],dist[u][0]);
            dist[u][0]=dfs1(v,u)+edges[i].w;
        }
        else if(dist[u][1]<dfs1(v,u)+edges[i].w)
            dist[u][1]=max(dist[u][1],dfs1(v,u)+edges[i].w);
    }
    return dist[u][0];
}

void dfs2(int u,int fa) {
    for(int i=head[u];i!=-1;i=edges[i].next) {
        int v = edges[i].to;
        if(v==fa)continue;
        if(v==longest[u]) dist[v][2]=max(dist[u][2],dist[u][1])+edges[i].w;
        else dist[v][2]=max(dist[u][2],dist[u][0])+edges[i].w;
        dfs2(v,u);
    }
}

int finish(int n,int M){
    if(n <= 0) return n;
    if(M < 0) return 0;
    deque<int> qmax,qmin;           //qmax单调递减 qmin单调递增
    deque<int> idmax,idmin;         //id存节点编号(同样单调)
    int ans=0;
    int left=1,right=1;
    while(right <= n){
    //维护单调性
		while(!qmax.empty() && dis[right] >= qmax.back()) qmax.pop_back(), idmax.pop_back();
		qmax.push_back(dis[right]); idmax.push_back(right); 
		while(!qmin.empty() && dis[right] <= qmin.back()) qmin.pop_back(), idmin.pop_back();
		qmin.push_back(dis[right]); idmin.push_back(right);
		while(qmax.front()-qmin.front() > M && left<right){  //超出m就减少最大值/增大最小值(左指针右移)
			left++;                  
			while(idmax.front() < left) idmax.pop_front(), qmax.pop_front();
			while(idmin.front() < left) idmin.pop_front(), qmin.pop_front();
		}
		ans = max(ans,right-left+1);
		right++;
	}
		return ans;
}
int main() {
    int n,m;
    while(scanf("%d%d",&n,&m)==2&&n) {
        tot=0;
        memset(dist,-1,sizeof(dist));
        memset(head,-1,sizeof(head));
        memset(longest,-1,sizeof(longest));
        for(int i=2; i<=n; i++) {
            int v,w;
            scanf("%d%d",&v,&w);
            add_edge(i,v,w);
            add_edge(v,i,w);
        }
        dfs1(1,-1);
        dfs2(1,-1);
        for(int i=1;i<=n;i++) dis[i]=max(dist[i][0],dist[i][2]);
        printf("%d\n",finish(n,m));
    }
    return 0;
}

 

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