首页 > ACM题库 > HDU-杭电 > hdu 2586 How far away ?-动态规划-[解题报告]C++
2014
02-10

hdu 2586 How far away ?-动态规划-[解题报告]C++

How far away ?

问题描述 :

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can’t visit a place twice) between every two houses. Yout task is to answer all these curious people.

输入:

First line is a single integer T(T<=10), indicating the number of test cases.
  For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
  Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

输出:

First line is a single integer T(T<=10), indicating the number of test cases.
  For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
  Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

样例输入:

2
3 2
1 2 10
3 1 15
1 2
2 3

2 2
1 2 100
1 2
2 1

样例输出:

10
25
100
100

LCA模板题

 题意:给一个无根树,有q个询问,每个询问两个点,问两点的距离。求出  lca = LCA(X,Y) , 然后  dir[x] + dir[y] – 2 * dir[lca]

dir[u]表示点u到树根的距离

 

下面两份代码都可以通过HDU的C++和G++,都不存在爆栈问题,网上很多人说会爆栈,加了申请系统栈语句,其实不用,而且好想比赛中不允许使用的

Tarjan算法跑得更快些,C++ 15ms,  G++ 50ms  左右, RMQ大概60ms

 

在线算法:LCA转化为RMQ

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
//#pragma comment(linker, "/STACK:102400000,102400000") //不需要申请系统栈
const int N = 40010;
const int M = 25;

int _pow[M];
int tot,head[N],ver[2*N],R[2*N],first[N],dir[N];
int dp[2*N][M];  //这个数组记得开到2*N,因为遍历后序列长度为2*n-1
bool vis[N];
struct edge
{
    int u,v,w,next;
}e[2*N];

inline void add(int u ,int v ,int w ,int &k)
{
    e[k].u = u; e[k].v = v; e[k].w = w; 
    e[k].next = head[u]; head[u] = k++;
    u = u^v; v = u^v; u = u^v;
    e[k].u = u; e[k].v = v; e[k].w = w; 
    e[k].next = head[u]; head[u] = k++;
}

void dfs(int u ,int dep)
{
    vis[u] = true; ver[++tot] = u; first[u] = tot; R[tot] = dep;
    for(int k=head[u]; k!=-1; k=e[k].next)
        if( !vis[e[k].v] )
        {
            int v = e[k].v , w = e[k].w;
            dir[v] = dir[u] + w;
            dfs(v,dep+1);
            ver[++tot] = u; R[tot] = dep;
        }
}

void ST(int len)
{
    int K = (int)(log((double)len) / log(2.0));
    for(int i=1; i<=len; i++) dp[i][0] = i;
    for(int j=1; j<=K; j++)
        for(int i=1; i+_pow[j]-1<=len; i++)
        {
            int a = dp[i][j-1] , b = dp[i+_pow[j-1]][j-1];
            if(R[a] < R[b]) dp[i][j] = a;
            else            dp[i][j] = b;
        }
}

int RMQ(int x ,int y)
{
    int K = (int)(log((double)(y-x+1)) / log(2.0));
    int a = dp[x][K] , b = dp[y-_pow[K]+1][K];
    if(R[a] < R[b]) return a;
    else            return b;
}

int LCA(int u ,int v)
{
    int x = first[u] , y = first[v];
    if(x > y) swap(x,y);
    int res = RMQ(x,y);
    return ver[res];
}

int main()
{
    for(int i=0; i<M; i++) _pow[i] = (1<<i);
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        int n,q,num = 0;
        scanf("%d%d",&n,&q);
        memset(head,-1,sizeof(head));
        memset(vis,false,sizeof(vis));
        for(int i=1; i<n; i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w,num);
        }
        tot = 0; dir[1] = 0;
        dfs(1,1);
        /*
        printf("节点 "); for(int i=1; i<=2*n-1; i++) printf("%d ",ver[i]); cout << endl;
        printf("深度 "); for(int i=1; i<=2*n-1; i++) printf("%d ",R[i]);   cout << endl;
        printf("首位 "); for(int i=0; i<n; i++) printf("%d ",first[i]);    cout << endl;
        printf("距离 "); for(int i=0; i<n; i++) printf("%d ",dir[i]);      cout << endl;
        */
        ST(2*n-1);
        while(q--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            int lca = LCA(u,v);
//            printf("lca = %d\n",lca);
            printf("%d\n",dir[u] + dir[v] - 2*dir[lca]);
        }
    }
    return 0;
}

 

 

离线算法:Tarjan算法解决

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 40010;
const int M = 410;

int head[N],__head[N];
struct edge{
    int u,v,w,next;
}e[2*N];
struct ask{
    int u,v,lca,next;
}ea[M];
int dir[N],fa[N],ance[N];
bool vis[N];

inline void add_edge(int u,int v,int w,int &k)
{
    e[k].u = u; e[k].v = v; e[k].w = w;
    e[k].next = head[u]; head[u] = k++;
    u = u^v; v = u^v; u = u^v;
    e[k].u = u; e[k].v = v; e[k].w = w;
    e[k].next = head[u]; head[u] = k++;
}

inline void add_ask(int u ,int v ,int &k)
{
    ea[k].u = u; ea[k].v = v; ea[k].lca = -1;
    ea[k].next = __head[u]; __head[u] = k++;
    u = u^v; v = u^v; u = u^v;
    ea[k].u = u; ea[k].v = v; ea[k].lca = -1;
    ea[k].next = __head[u]; __head[u] = k++;
}

int Find(int x)
{
    return x == fa[x] ? x : fa[x] = Find(fa[x]);
}
void Union(int u ,int v)
{
    fa[v] = fa[u];  //可写为  fa[Find(v)] = fa[u];
}

void Tarjan(int u)
{
    vis[u] = true;
    ance[u] = fa[u] = u; //课写为 ance[Find(u)] = fa[u] = u;
    for(int k=head[u]; k!=-1; k=e[k].next)
        if( !vis[e[k].v] )
        {
            int v = e[k].v , w = e[k].w;
            dir[v] = dir[u] + w;
            Tarjan(v);
            Union(u,v);
            //ance[Find(u)] = u;  //可写为ance[u] = u;  //甚至不要这个语句都行
        }
    for(int k=__head[u]; k!=-1; k=ea[k].next)
        if( vis[ea[k].v] )
        {
            int v = ea[k].v;
            ea[k].lca = ea[k^1].lca = ance[Find(v)];
        }
}

int main()
{
    int cas,n,q,tot;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%d%d",&n,&q);
        memset(head,-1,sizeof(head));
        memset(__head,-1,sizeof(__head));
        tot = 0;
        for(int i=1; i<n; i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            add_edge(u,v,w,tot);
        }
        tot = 0;
        for(int i=0; i<q; i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            add_ask(u,v,tot);
        }
        memset(vis,0,sizeof(vis));
        dir[1] = 0;
        Tarjan(1);
        for(int i=0; i<q; i++)
        {
            int s = i * 2 , u = ea[s].u , v = ea[s].v , lca = ea[s].lca;
            printf("%d\n",dir[u] + dir[v] - 2*dir[lca]);
        }
    }
    return 0;
}

 

解题转自:http://www.cnblogs.com/scau20110726/archive/2013/05/26/3100265.html