How far away ？
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5492 Accepted Submission(s): 2090
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.
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.
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
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
25
100
100
Source
Recommend
用邻接表+dfs比较容易过...
代码：
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn=;
struct node
{
int id,val;
};
bool vis[maxn];
vector< node >map[maxn];
node tem;
int n,m,ans,cnt;
void dfs(int a,int b)
{ if(a==b){
if(ans>cnt)ans=cnt;
return ;
}
int Size=map[a].size();
vis[a]=;
for(int i=;i<Size;i++){
if(!vis[map[a][i].id]){
cnt+=map[a][i].val;
dfs(map[a][i].id,b);
cnt-=map[a][i].val;
}
}
vis[a]=;
}
int main()
{
int cas,a,b,val;
cin>>cas;
while(cas--){
cin>>n>>m;
cnt=;
for(int i=;i<=n;i++)
map[i].clear();
for(int i=;i<n;i++){
scanf("%d%d%d",&a,&b,&val); tem=(node){b,val};
map[a].push_back(tem); //ÎÞÏòͼ
tem=(node){a,val};
map[b].push_back(tem);
}
for(int i=;i<m;i++)
{
ans=0x3f3f3f3f;
scanf("%d%d",&a,&b);
dfs(a,b);
printf("%d\n",ans);
}
}
return ;
}