clj在某场hihoCoder比赛中的一道题,表示clj的数学题实在6,这道图论貌似还算可以。。。
题目链接:http://hihocoder.com/problemset/problem/1167
由于是中文题目,题意不再赘述。
对于任意两条小精灵的活动路径a和b,二者相交的判断条件为b的两个端点的LCA在a的路径上;那么我们可以首先将每个活动路径端点的LCA离线预处理出来,对每个节点LCA值+1。
然后以某个节点(我选择的是节点1)为根进行深搜,算出一条从节点1到节点x的LCA值和,那么任意路径a(假设其两端点分别是A和B)上的节点个数就是sum[A] + sum[B] - 2 * sum[LCA(A,B)]。
最后,对于某些点,如果它是不止一条路径的LCA,那么我们只需要对最终答案乘以C(LCAnum, 2)的组合数就好。
【PS:clj给出的题解中,采用了点分治+LCA的方式,虽然看懂了题意,但是表示对递归分治之后的路径,如何求出其上的LCAnum,并没有多好的想法,还望巨巨能指点一下,Thx~】
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 using namespace std; 5 typedef long long LL; 6 #define MAXN 100010 7 struct Edge { 8 int to, next; 9 } edge[MAXN << 1]; 10 struct Node { 11 int to, next, num; 12 } Query[MAXN << 1]; 13 struct node { 14 int u, v, lca; 15 } input[MAXN]; 16 int totEdge, totQuery, n, m; 17 int headEdge[MAXN], headQuery[MAXN]; 18 int ancestor[MAXN], father[MAXN], LCAnum[MAXN], sum[MAXN]; 19 bool vis[MAXN]; 20 void addEdge(int from, int to) { 21 edge[totEdge].to = to; 22 edge[totEdge].next = headEdge[from]; 23 headEdge[from] = totEdge++; 24 } 25 void addQuery(int from, int to, int x) { 26 Query[totQuery].to = to; 27 Query[totQuery].num = x; 28 Query[totQuery].next = headQuery[from]; 29 headQuery[from] = totQuery++; 30 } 31 void init() { 32 memset(headEdge, -1, sizeof(headEdge)); 33 memset(headQuery, -1, sizeof(headQuery)); 34 memset(father, -1, sizeof(father)); 35 memset(vis, false, sizeof(vis)); 36 memset(sum, 0, sizeof(sum)); 37 memset(LCAnum, 0, sizeof(LCAnum)); 38 totEdge = totQuery = 0; 39 } 40 int find_set(int x) { 41 if(x == father[x]) return x; 42 else return father[x] = find_set(father[x]); 43 } 44 void union_set(int x, int y) { 45 x = find_set(x); y = find_set(y); 46 if(x != y) father[y] = x; 47 } 48 void Tarjan(int u) { 49 father[u] = u; 50 for(int i = headEdge[u]; i != -1; i = edge[i].next) { 51 int v = edge[i].to; 52 if(father[v] != -1) continue; 53 Tarjan(v); 54 union_set(u, v); 55 } 56 for(int i = headQuery[u]; i != -1; i = Query[i].next) { 57 int v = Query[i].to; 58 if(father[v] == -1) continue; 59 input[Query[i].num].lca = find_set(v); 60 } 61 } 62 void DFS(int u, int pre) { 63 vis[u] = 1; 64 sum[u] = sum[pre] + LCAnum[u]; 65 for(int i = headEdge[u]; i != -1; i = edge[i].next) { 66 int v = edge[i].to; 67 if(vis[v]) continue; 68 DFS(v, u); 69 } 70 } 71 int main() { 72 init(); 73 scanf("%d%d", &n, &m); 74 for(int i = 0; i < n - 1; i++) { 75 int a, b; 76 scanf("%d%d", &a, &b); 77 addEdge(a, b); addEdge(b, a); 78 } 79 for(int i = 0; i < m; i++) { 80 int a, b; 81 scanf("%d%d", &a, &b); 82 input[i].u = a, input[i].v = b; 83 addQuery(a, b, i); addQuery(b, a, i); 84 } 85 Tarjan(1); 86 for(int i = 0; i < m; i++) 87 LCAnum[input[i].lca]++; 88 DFS(1, 0); 89 LL ans = 0; 90 for(int i = 0; i < m; i++) { 91 ans += (sum[input[i].u] + sum[input[i].v] - 2 * sum[input[i].lca]); 92 } 93 for(int i = 1; i <= n; i++) { 94 ans += (LL)LCAnum[i] * (LCAnum[i] - 1) / 2; 95 } 96 printf("%lld ", ans); 97 return 0; 98 }
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