HDU 4123 Bob’s Race 树形dp+单调队列

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4123

Time Limit: 5000/2000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others) #### 问题描述 > 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. #### 输入 > 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) #### 输出 > For each test case, you should output the answer in a line for each query. ####样例输入 > 5 5 > 1 2 3 > 2 3 4 > 4 5 3 > 3 4 2 > 1 > 2 > 3 > 4 > 5 > 0 0

样例输出

1
3
3
3
5

题意

给你一颗树，如果设置的起点包含点v，那么从那里出发的选手就会跑到离它最远的顶点上。现在每个查询给你一个q，让你选最多的起点，同时保证所有选手中跑的最大距离和最小距离要小于等于q。

题解

先用树形dp处理直径的方式处理出每个点能跑的最远距离。然后就是一个裸的单调队列的问题了。（单调队列是可以维护区间最值的），当然，你用rmq或线段树维护也是可以的。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=50505;

int dp[maxn][2],id[maxn];
int arr[maxn];
int n,m;
VPII G[maxn];

///单调队列，用优先队列的t了，手写了一个O(n)的。
struct Queue{
    int lis[maxn],f,r;
    int clk[maxn];
    void init(){ f=1,r=0; }
    bool empty(){ return f>r; }
    void pop_front(){ f++; }
    void pop_back(){ r--; }
    void push(int v,int t){
        lis[++r]=v;
        clk[r]=t;
    }
    int front(int istim){
        if(empty()) return -1;
        if(istim) return clk[f];
        return lis[f];
    }
    int rear(int istim){
        if(istim) return clk[r];
        return lis[r];
    }
}q1,q2;

///树形dp求每个点的最远顶点距离
void dfs(int u,int fa){
    id[u]=-1;
    dp[u][0]=dp[u][1]=0;
    rep(i,0,G[u].sz()){
        int v=G[u][i].X,w=G[u][i].Y;
        if(v==fa) continue;
        dfs(v,u);
        ///更新最大值
        if(dp[u][0]<dp[v][0]+w){
            dp[u][1]=dp[u][0];
            ///更新次大
            dp[u][0]=dp[v][0]+w;
            id[u]=v;
        }else if(dp[u][1]<dp[v][0]+w){
            ///更新次大
            dp[u][1]=dp[v][0]+w;
        }
    }
}

///用父亲下来的最远距离更新最值
void dfs2(int u,int fa,int ma){
    dp[u][0]=max(dp[u][0],ma);
    rep(i,0,G[u].sz()){
        int v=G[u][i].X,w=G[u][i].Y;
        if(v==fa) continue;
        if(id[u]==v){
            dfs2(v,u,max(dp[u][1],ma)+w);
        }else{
            dfs2(v,u,max(dp[u][0],ma)+w);
        }
    }
}

void init(){
    for(int i=1;i<=n;i++) G[i].clear();
}

int main() {
    while(scf("%d%d",&n,&m)==2&&n){
        init();
        rep(i,0,n-1){
            int u,v,w;
            scf("%d%d%d",&u,&v,&w);
            G[u].pb(mkp(v,w));
            G[v].pb(mkp(u,w));
        }

        dfs(1,-1);
        dfs2(1,-1,0);

        for(int i=1;i<=n;i++) arr[i]=dp[i][0];

        while(m--){
            int q; scf("%d",&q);

            ///q1小,q2大,单调队列
            q1.init(),q2.init();

            int l=1,Ma=1;

            for(int i=1;i<=n;i++){
                while(!q1.empty()&&q1.rear(0)>=arr[i]) q1.pop_back();
                while(!q2.empty()&&q2.rear(0)<=arr[i]) q2.pop_back();
                q1.push(arr[i],i);
                q2.push(arr[i],i);

                while(q2.front(0)-q1.front(0)>q){
                    l++;
                    while(q1.front(1)<l) q1.pop_front();
                    while(q2.front(1)<l) q2.pop_front();
                }

                Ma=max(Ma,i-l+1);
            }

            prf("%d\n",Ma);

        }

    }
    return 0;
}

/*
5 3
1 2 9
3 4 5
3 5 3
*/