sizeof rmq round pad show mod 部分 cas sam

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=2586

Problem Description 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.
Input 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.
Output For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.
Sample Input

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

2 2
1 2 100
1 2
2 1

Sample Output

10
25
100
100

Source ECJTU 2009 Spring Contest

題意：

一個村莊有 n 個房子和 n-1 條雙向路，每兩個房子之間都有一條簡單路徑。

如今有m次詢問。求兩房子之間的距離。

PS:

能夠用LCA來解，首先找到u, v 兩點的lca，然後計算一下距離值就能夠了。

計算方法是。記下根結點到隨意一點的距離dis[i]，

這樣ans = dis[u] + dis[v] - 2 * dis[lca(v, v)]了。

這題要用c++交。G++會爆棧！

代碼例如以下：看別人的模板（tarjan 離線）

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 40047
#define maxm 247

struct node
{
    int to,w,next;
} edge[maxn*2];

int n, m;		//點數，詢問次數
int head[maxn];
int k;
int fa[maxn];		//父親結點
int dis[maxn];		//到根節點距離
int vis[maxn];		//是否訪問過
int s[maxm];		//詢問起點
int e[maxm];		//詢問終點
int lca[maxm];		//LCA(s,e) 近期公共祖先

int find(int x)
{
    if(fa[x]!=x) return fa[x]=find(fa[x]);
    return fa[x];
}

void init()
{
    k = 1;
    memset(head,0,sizeof(head));
    memset(dis,0,sizeof(dis));
    memset(vis,0,sizeof(vis));
}

void add(int u,int v,int w)
{
    edge[k].to = v;
    edge[k].w = w;
    edge[k].next = head[u];
    head[u] = k++;
}

void tarjan(int u)
{
    int i,v;
    fa[u] = u;
    vis[u] = 1;
    for(i = 0; i < m; i++)
    {
        if(e[i]==u && vis[s[i]])
            lca[i] = find(s[i]);	//若詢問的兩點中有一點已被訪問過。則兩點的LCA則為這一點的當前父節點
        if(s[i]==u && vis[e[i]])
            lca[i] = find(e[i]);
    }
    for(i = head[u]; i; i = edge[i].next)
    {
        v = edge[i].to;
        if(!vis[v]) //若沒被訪問過
        {
            dis[v] = dis[u]+edge[i].w;//更新距離
            tarjan(v);
            fa[v] = u;//回溯更新父節點
        }
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        scanf("%d%d",&n,&m);
        int u, v, w;
        for(int i = 0; i < n-1; i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d",&s[i],&e[i]);
        }
        tarjan(1);

        for(int i = 0; i < m; i++)
        {
            printf("%d\n",dis[s[i]]+dis[e[i]]-2*dis[lca[i]]);//兩點距離為根節點到兩點距離之和-根節點到LCA距離*2
        }
    }
    return 0;
}

（ST在線算法 轉）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
//#pragma comment(linker, "/STACK:102400000,102400000") //不須要申請系統棧
const int N = 40010;
const int M = 25;
int dp[2*N][M];  //這個數組記得開到2*N，由於遍歷後序列長度為2*n-1
bool vis[N];
struct edge
{
    int u,v,w,next;
} e[2*N];
int tot,head[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++;
}
int ver[2*N],R[2*N],first[N],dir[N];
//ver:節點編號 R：深度 first：點編號位置 dir：距離
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 n)
{
    for(int i=1; i<=n; i++)
        dp[i][0] = i;
    for(int j=1; (1<<j)<=n; j++)
    {
        for(int i=1; i+(1<<j)-1<=n; i++)
        {
            int a = dp[i][j-1] , b = dp[i+(1<<(j-1))][j-1];
            dp[i][j] = R[a]<R[b]?a:b;
        }
    }
}
//中間部分是交叉的。
int RMQ(int l,int r)
{
    int k=0;
    while((1<<(k+1))<=r-l+1)
        k++;
    int a = dp[l][k], b = dp[r-(1<<k)+1][k]; //保存的是編號
    return R[a]<R[b]?a: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()
{
    //freopen("Input.txt","r",stdin);
    //freopen("Out.txt","w",stdout);
    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("節點ver "); for(int i=1; i<=2*n-1; i++) printf("%d ",ver[i]); cout << endl;
        printf("深度R "); for(int i=1; i<=2*n-1; i++) printf("%d ",R[i]);   cout << endl;
        printf("首位first "); for(int i=1; i<=n; i++) printf("%d ",first[i]);    cout << endl;
        printf("距離dir "); for(int i=1; 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("%d\n",dir[u] + dir[v] - 2*dir[lca]);
        }
    }
    return 0;
}

HDU 2586 How far away ？（LCA模板 近期公共祖先啊）