hdu----(2586)How far away ?(DFS/LCA/RMQ)
2014-09-05 14:17
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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
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.
[align=left]Sample Input[/align]
2
3 2
1 2 10
3 1 15
1 2
2 3
2 2
1 2 100
1 2
2 1
[align=left]Sample Output[/align]
10
25
100
100
[align=left]Source[/align]
ECJTU 2009 Spring Contest
[align=left]Recommend[/align]
[align=left] [/align]
[align=left]用邻接表+dfs比较容易过...[/align]
[align=left]代码:[/align]
#include<cstring> #include<cstdio> #include<cstdlib> #include<vector> #include<algorithm> #include<iostream> using namespace std; const int maxn=40100; 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]=1; for(int i=0;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]=0; } int main() { int cas,a,b,val; cin>>cas; while(cas--){ cin>>n>>m; cnt=0; for(int i=1;i<=n;i++) map[i].clear(); for(int i=1;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=0;i<m;i++) { ans=0x3f3f3f3f; scanf("%d%d",&a,&b); dfs(a,b); printf("%d\n",ans); } } return 0; }
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