while nts 分享 put numbers ont integer simple people

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 ECJTU 2009 Spring Contest 題目大意： 共有t組數據 給你n個點，n-1條雙向邊 有m次詢問 問x~y點之間的距離是多少

 1 #include <cstring>
 2 #include <ctype.h>
 3 #include <cstdio>
 4 
 5 const int MAXN=40010;
 6 
 7 int t,n,m;
 8 
 9 struct node {
10     int to;
11     int next;
12     int val;
13 };
14 node e[MAXN<<1];
15 
16 int head[MAXN],tot;
17 
18 int deep[MAXN],f[MAXN][21],dis[MAXN];
19 
20 inline void read(int&x) {
21     int f=1;register char c=getchar();
22     for(x=0;!isdigit(c);c==‘-‘&&(f=-1),c=getchar());
23     for(;isdigit(c);x=x*10+c-48,c=getchar());
24     x=x*f;
25 }
26 
27 inline void add(int x,int y,int v) {
28     e[++tot].to=y;
29     e[tot].val=v;
30     e[tot].next=head[x];
31     head[x]=tot;
32 }
33 
34 void dfs(int u) {
35     deep[u]=deep[f[u][0]]+1;
36     for(int i=head[u];i!=-1;i=e[i].next) {
37         int to=e[i].to;
38         if(!deep[to]&&to) {
39             f[to][0]=u;
40             dis[to]=dis[u]+e[i].val;
41             dfs(to);
42         }
43     }
44     return;
45 }
46 
47 inline void swap(int&x,int&y) {
48     int t=x;
49     x=y;y=t;
50     return;
51 }
52 
53 inline int LCA(int x,int y) {
54     if(deep[x]<deep[y]) swap(x,y);
55     int t=deep[x]-deep[y];
56     for(int i=20;i>=0;--i) 
57       if(deep[f[x][i]]>=deep[y]) x=f[x][i];
58     if(x==y) return x;
59     for(int i=20;i;--i) 
60       if(f[x][i]!=f[y][i])
61         x=f[x][i],y=f[y][i];
62     return f[x][0];
63     
64 }
65 
66 inline void pre() {
67     tot=1;
68     memset(head,-1,sizeof head);    
69     memset(deep,0,sizeof deep);
70     memset(dis,0,sizeof dis);
71 }
72 
73 int hh() {
74     int x,y,z;
75     read(t);
76     while(t--) {
77         read(n);read(m);
78         pre();
79         for(int i=1;i<n;++i) {
80             read(x);read(y);read(z);
81             add(x,y,z);
82             add(y,x,z);
83         }
84         dfs(1);
85         for(int j=1;j<=20;++j) 
86           for(int i=1;i<=n;++i) 
87             f[i][j]=f[f[i][j-1]][j-1];
88         for(int i=1;i<=m;++i) {
89             read(x);read(y);
90             int lca=LCA(x,y);
91             printf("%d\n",dis[x]+dis[y]-2*dis[lca]);
92         }
93     }
94     return 0;
95 }
96 
97 int sb=hh();
98 int main() {;}

代碼

HDU How far away ？--LCA