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
Source
思路:在lca的基础上加dis(距离根的距离)数组,在dfs处理好,ans=dis[a]-2*dis[LCA(a,b)]+dis[b];
#include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> #define true ture #define false flase using namespace std; #define ll long long #define inf 0xfffffff int scan() { int res = 0 , ch ; while( !( ( ch = getchar() ) >= ‘0‘ && ch <= ‘9‘ ) ) { if( ch == EOF ) return 1 << 30 ; } res = ch - ‘0‘ ; while( ( ch = getchar() ) >= ‘0‘ && ch <= ‘9‘ ) res = res * 10 + ( ch - ‘0‘ ) ; return res ; } #define maxn 40010 #define M 22 struct is { int v,next,w; } edge[maxn*2]; int deep[maxn],jiedge; int dis[maxn]; int head[maxn]; int rudu[maxn]; int fa[maxn][M]; void add(int u,int v,int w) { jiedge++; edge[jiedge].v=v; edge[jiedge].w=w; edge[jiedge].next=head[u]; head[u]=jiedge; } void dfs(int u) { for(int i=head[u]; i; i=edge[i].next) { int v=edge[i].v; int w=edge[i].w; if(!deep[v]) { dis[v]=dis[u]+edge[i].w; deep[v]=deep[u]+1; fa[v][0]=u; dfs(v); } } } void st(int n) { for(int j=1; j<M; j++) for(int i=1; i<=n; i++) fa[i][j]=fa[fa[i][j-1]][j-1]; } int LCA(int u , int v) { if(deep[u] < deep[v]) swap(u , v) ; int d = deep[u] - deep[v] ; int i ; for(i = 0 ; i < M ; i ++) { if( (1 << i) & d ) // 注意此处,动手模拟一下,就会明白的 { u = fa[u][i] ; } } if(u == v) return u ; for(i = M - 1 ; i >= 0 ; i --) { if(fa[u][i] != fa[v][i]) { u = fa[u][i] ; v = fa[v][i] ; } } u = fa[u][0] ; return u ; } void init() { memset(head,0,sizeof(head)); memset(fa,0,sizeof(fa)); memset(rudu,0,sizeof(rudu)); memset(deep,0,sizeof(deep)); jiedge=0; } int main() { int x,n; int t; scanf("%d",&t); while(t--) { init(); scanf("%d%d",&n,&x); for(int i=1; i<n; i++) { int u,v,w; scanf("%d%d%d",&u,&v,&w); add(u,v,w); rudu[v]++; } for(int i=1;i<=n;i++) { if(!rudu[i]) { deep[i]=1; dis[i]=0; dfs(i); break; } } st(n); while(x--) { int a,b; scanf("%d%d",&a,&b); printf("%d\n",dis[a]-2*dis[LCA(a,b)]+dis[b]); } } return 0; }