UESTC - 92 LCA 倍增

Bob has traveled to byteland, he find the  N N cities in byteland formed a tree structure, a tree structure 
is very special structure, there is exactly one path connecting each pair of nodes, and a tree with  N N 
nodes has  N−1 N−1 edges.

As a traveler, Bob wants to journey between those  N N cities, and he know the time each road will cost. 
he advises the king of byteland building a new road to save time, and then, a new road was built. Now Bob 
has  Q Q journey plan, give you the start city and destination city, please tell Bob how many time is saved 
by add a road if he always choose the shortest path. Note that if it's better not journey from the new roads, 
the answer is  0 0.

First line of the input is a single integer  T T( 1≤T≤20 1≤T≤20), indicating there are  T T test cases.

For each test case, the first will line contain two integers  N N( 2≤N≤105 2≤N≤105) and  Q Q( 1≤Q≤105 1≤Q≤105), indicating 
the number of cities in byteland and the journey plans. Then  N N line followed, each line will contain three 
integer  x x,  y y( 1≤x,y≤N 1≤x,y≤N) and  z z( 1≤z≤1000 1≤z≤1000) indicating there is a road cost  z z time connect 
the  xth xth city and the  yth yth city, the first  N−1 N−1 roads will form a tree structure, indicating the original roads, 
and the  Nth Nth line is the road built after Bob advised the king. Then  Q Qline followed, each line will contain two 
integer  x x and  y y( 1≤x,y≤N 1≤x,y≤N), indicating there is a journey plan from the  xth xth city to  yth yth city.

For each case, you should first output Case #t: in a single line, where  t t indicating the case number between  1 1 and  T T, 
then  Q Q lines followed, the  ith ith line contains one integer indicating the time could saved in  ith ith journey plan.

1
5 5
1 2 3
2 3 4
4 1 5
3 5 1
3 1 5
1 2
1 3
2 5
3 4
4 5

Case #1:
0
2
0
2
2


题意：在原有图的基础上加一条边。问加边之后是不是减少了距离，如果减少了就输出减少了多少。
如果没减少就输出0.
LCA水题



#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 = 1e5+100;
const int M = 35;
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];
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:节点编号 R：深度 first：点编号位置 dir：距离
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;
        }
    }
}
//中间部分是交叉的。
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]; //保存的是编号
    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);
    int cnt=ver[res];
    return dir[u]+dir[v]-2*dir[cnt];
}



int main()
{
    //freopen("Input.txt","r",stdin);
    //freopen("Out.txt","w",stdout);
    int t;
    scanf("%d",&t);
    for(int cas=1;cas<=t;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);
        }
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        printf("Case #%d:n",cas);
        tot = 0; dir[1] = 0;
        dfs(1,1);
        ST(2*n-1);
        while(q--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            int lca = LCA(u,v);//u v 之间的距离
           // cout<<lca<<"----------"<<endl;
            int lca1=min(LCA(u,a)+LCA(v,b),LCA(u,b)+LCA(v,a))+c;
            lca-=lca1;
            //cout<<lca1<<"*********"<<endl;
            if(lca<0) lca=0;
            printf("%dn",lca);
        }
    }
}