hdu 2586 LCA

链接：戳这里

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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
 

题意：

n个点的树，边带权值，m组询问u,v。输出u到v之间的路径权值总和

思路：

LCA裸的题

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<vector>#include <ctime>#include<queue>#include<set>#include<map>#include<stack>#include<iomanip>#include<cmath>#include<bitset>#define mst(ss,b) memset((ss),(b),sizeof(ss))///#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef long double lb;#define INF (1ll<<60)-1#define Max 1e9using namespace std;int T;int n,m;ll f[40010][22];int fa[40010][22],deep[40010];int head[40010],tot;struct edge{    int v,next,w;}e[80010];void Add(int u,int v,int w){    e[tot].v=v;    e[tot].w=w;    e[tot].next=head[u];    head[u]=tot++;}void DFS(int u,int Fa,int de){    deep[u]=de;    fa[u][0]=Fa;    for(int i=head[u];i!=-1;i=e[i].next){        int v=e[i].v;        if(v==Fa) continue;        f[v][0]=(ll)e[i].w;        DFS(v,u,de+1);    }}void build(){    for(int i=1;i<=20;i++){        for(int j=1;j<=n;j++){            if(fa[j][i-1]!=-1){                fa[j][i]=fa[fa[j][i-1]][i-1];                f[j][i]=f[j][i-1]+f[fa[j][i-1]][i-1];            }        }    }}int LCA(int u,int v){    if(deep[u]<deep[v]) swap(u,v);    int len=deep[u]-deep[v];    for(int i=0;i<=20;i++){        if(len&(1<<i)) u=fa[u][i];    }    for(int i=20;i>=0;i--){        if(fa[u][i]!=fa[v][i]){            u=fa[u][i];            v=fa[v][i];        }    }    if(u==v) return u;    return fa[u][0];}int main(){    scanf("%d",&T);    while(T--){        mst(head,-1);tot=0;mst(deep,0);mst(fa,-1);mst(f,0);        scanf("%d%d",&n,&m);        for(int i=1;i<n;i++){            int u,v,w;            scanf("%d%d%d",&u,&v,&w);            Add(u,v,w);            Add(v,u,w);        }        DFS(1,0,0);        build();        while(m--){            int u,v;            scanf("%d%d",&u,&v);            int lca=LCA(u,v);            ///printf("lca=%d\n",lca);            ll ans=0;            int lenu=deep[u]-deep[lca];            for(int i=20;i>=0;i--){                if(lenu&(1<<i)) {                    ans+=f[u][i];                    u=fa[u][i];                }            }            int lenv=deep[v]-deep[lca];            for(int i=20;i>=0;i--){                if(lenv&(1<<i)) {                    ans+=f[v][i];                    v=fa[v][i];                }            }            printf("%I64d\n",ans);        }    }    return 0;}