83131

2015-12-29 20:44

这道题不难，就是3个点的lca。算法有点多，写成树链剖分的吧！跑完2400多毫秒。还好，挺顺利的，加油！努力啊！注意看数据范围！相信自己，能行的

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #define rep(i,j,k) for(int i = j; i <= k; i++)
 5 #define down(i,j,k) for(int i = j; i >= k; i--)
 6 #define maxn 500005
 7 using namespace std;
 8 
 9 struct edge{
10 int to; edge* next;
11 };
12 edge* head[500005], *pt, edges[1000005];
13 
14 void add_edge(int x,int y)
15 {
16     pt->to = x, pt->next = head[y], head[y] = pt++;
17     pt->to = y, pt->next = head[x], head[x] = pt++;
18 }
19 
20 inline int read()
21 {
22     int s = 0, t = 1; char c = getchar();
23     while( !isdigit(c) ){
24         if( c == '-' ) t = -1; c = getchar();
25     }
26     while( isdigit(c) ){
27         s = s * 10 + c - '0'; c = getchar();
28     }
29     return s * t;
30 }
31 
32 int top[maxn], dep[maxn], pa[maxn], size[maxn], son[maxn];
33 
34 void dfs(int now)
35 {
36     son[now] = 0, size[now] = 1;
37     for(edge*i = head[now]; i; i=i->next){
38         int to = i->to; if( to == pa[now] ) continue;
39         pa[to] = now, dep[to] = dep[now] + 1;
40         dfs(to);
41         if( size[to] > size[son[now]] ) son[now] = to;
42     }
43 }
44 
45 void dfs2(int now,int ph)
46 {
47     top[now] = ph;
48     if( son[now] ) dfs2(son[now],ph);
49     for(edge*i = head[now]; i; i=i->next){
50         int to = i->to; if( to == pa[now] || to == son[now] ) continue;
51         dfs2(to,to);
52     }
53 }
54 
55 int lca(int x,int y)
56 {
57     while( top[x] != top[y] ){
58         if( dep[top[x]] < dep[top[y]] ) swap(x,y);
59         x = pa[top[x]];
60     }
61     if(dep[x] < dep[y] ) return x;
62     else return y;
63 }
64 
65 int main()
66 {
67     int n = read(), m = read(); pt = edges; 
68     rep(i,1,n-1){
69         int x = read(), y = read();
70         add_edge(x,y);
71     }
72     dep[1] = 0; dfs(1); dfs2(1,1);
73     rep(i,1,m){
74         int x = read(), y = read(), z = read();
75         int lc1 = lca(x,y), lc2 = lca(y,z), lc3 = lca(x,z);
76         if( lc1 == lc2 ){
77             printf("%d %d\n",lc3,dep[y]-dep[lc1]+dep[x]-dep[lc1]+dep[z]-dep[lc3]);
78         }
79         else if( lc2 == lc3 ){
80             printf("%d %d\n", lc1,dep[z]-dep[lc2]+dep[y]-dep[lc2]+dep[x]-dep[lc1]);
81         }
82         else if( lc1 == lc3 ){
83             printf("%d %d\n", lc2,dep[x]-dep[lc1]+dep[y]-dep[lc1]+dep[z]-dep[lc2]);
84         }
85     }
86     return 0;
87 }

1787: [Ahoi2008]Meet 紧急集合

Time Limit: 20 Sec Memory Limit: 162 MB
Submit: 1829 Solved: 846
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Description

Input

Output

Sample Input

6 4
1 2
2 3
2 4
4 5
5 6
4 5 6
6 3 1
2 4 4
6 6 6

Sample Output

5 2
2 5
4 1
6 0

HINT

Source