How far away ？
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13641 Accepted Submission(s):
5117
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.
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
Source
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玛德这题为什么总是TLE……不是很懂>...<
1 #include "bits/stdc++.h" 2 #define mem(a,b) memset(a,b,sizeof(a)) 3 using namespace std; 4 typedef long long LL; 5 const int MAX1=40005; 6 const int MAX2=405; 7 int cas; 8 int n,m; 9 int tot,tq; 10 int head[MAX1],hq[MAX1]; 11 struct Edge{ int v,w; int next;} edge[MAX2],eq[MAX2]; 12 int ans[MAX2],dis[MAX1]; 13 int fa[MAX1]; 14 bool vis[MAX1]; 15 inline void addedge(int u,int v,int w) {tot++; edge[tot].v=v;edge[tot].w=w; edge[tot].next=head[u]; head[u]=tot;} 16 inline void addq(int u,int v) {tq++; eq[tq].v=v; eq[tq].next=hq[u]; hq[u]=tq;} 17 int getfather(int x) {if (fa[x]==x) return x;return fa[x]=getfather(fa[x]);} 18 void dfs(int x,int ff,int w){ 19 dis[x]=w; 20 for (int i=head[x];i;i=edge[i].next){ 21 if (edge[i].v==ff) continue; 22 dfs(edge[i].v,x,w+edge[i].w); 23 } 24 } 25 void init(){ 26 int i,j; 27 int u,v,w; 28 scanf("%d%d",&n,&m); 29 mem(ans,0),mem(vis,false),mem(head,0),mem(hq,0),mem(dis,0); 30 for (i=1;i<=n;i++) 31 fa[i]=i; 32 tot=tq=0; 33 for (i=1;i<n;i++){ 34 scanf("%d%d%d",&u,&v,&w); 35 addedge(u,v,w); 36 addedge(v,u,w); 37 } 38 for (i=1;i<=m;i++){ 39 scanf("%d%d",&u,&v); 40 addq(u,v); 41 } 42 } 43 void tarjanlca(int x){ 44 fa[x]=x;vis[x]=true; 45 int i,j; 46 for (i=head[x];i;i=edge[i].next){ 47 if (!vis[edge[i].v]){ 48 tarjanlca(edge[i].v); 49 fa[edge[i].v]=x; 50 } 51 } 52 for (i=hq[x];i;i=eq[i].next){ 53 if (vis[eq[i].v]){ 54 j=getfather(eq[i].v); 55 ans[i]=dis[x]+dis[eq[i].v]-2*dis[j]; 56 } 57 } 58 } 59 int main(){ 60 freopen ("away.in","r",stdin); 61 freopen ("away.out","w",stdout); 62 int i,j; 63 scanf("%d",&cas); 64 while (cas--){ 65 init(); 66 dfs(1,0,0); 67 tarjanlca(1); 68 for (i=1;i<=m;i++) 69 printf("%d\n",ans[i]); 70 } 71 return 0; 72 }