大二训练第二周 F - How far away ？ lca

F - How far away ？

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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

23 21 2 103 1 151 22 32 21 2 1001 22 1 

 

Sample Output

1025100100 

 

模板题啦。。

ACcode：

 

#pragma warning(disable:4786)//使命名长度不受限制#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈#include <map>#include <set>#include <queue>#include <cmath>#include <stack>#include <cctype>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define rds(x) scanf("%s",x)#define rdc(x) scanf("%c",&x)#define ll long long int#define maxn 100005#define mod 1000000007#define INF 0x3f3f3f3f //int 最大值#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define MT(x,i) memset(x,i,sizeof(x))#define PI  acos(-1.0)#define E  exp(1)using namespace std;struct node{    int id,val,next;}e[maxn<<1];int fa[maxn],dis[maxn],pre[maxn],x[maxn],y[maxn],z[maxn];bool vis[maxn];int n,m,cnt,loop;void add(int u,int v,int w){    e[cnt].id=u;    e[cnt].val=w;    e[cnt].next=pre[v];    pre[v]=cnt++;    e[cnt].id=v;    e[cnt].val=w;    e[cnt].next=pre[u];    pre[u]=cnt++;}int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}void tarjan(int k){    vis[k]=true;    fa[k]=k;    FOR(i,1,m){        if(x[i]==k&&vis[y[i]])            z[i]=find(y[i]);        if(y[i]==k&&vis[x[i]])            z[i]=find(x[i]);    }    for(int i=pre[k];i!=-1;i=e[i].next){        if(!vis[e[i].id]){            dis[e[i].id]=dis[k]+e[i].val;            tarjan(e[i].id);            fa[e[i].id]=k;        }    }}void init(){    FOR(i,0,maxn){        pre[i]=-1;        vis[i]=x[i]=y[i]=z[i]=0;    }}int main(){    rd(loop);    while(loop--){        int u,v,w;        rd2(n,m);cnt=0;        init();        FOR(i,1,n-1){            rd2(u,v);rd(w);            add(u,v,w);        }        FOR(i,1,m)rd2(x[i],y[i]);        dis[1]=0;tarjan(1);        FOR(i,1,m)            printf("%d\n",dis[x[i]]+dis[y[i]]-2*dis[z[i]]);    }    return 0;}/*23 21 2 103 1 151 22 32 21 2 1001 22 1*/