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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~】
AC代码:
#include <cstdio> #include <iostream> #include <cstring> using namespace std; typedef long long LL; #define MAXN 100010 struct Edge { int to, next; } edge[MAXN << 1]; struct Node { int to, next, num; } Query[MAXN << 1]; struct node { int u, v, lca; } input[MAXN]; int totEdge, totQuery, n, m; int headEdge[MAXN], headQuery[MAXN]; int ancestor[MAXN], father[MAXN], LCAnum[MAXN], sum[MAXN]; bool vis[MAXN]; void addEdge(int from, int to) { edge[totEdge].to = to; edge[totEdge].next = headEdge[from]; headEdge[from] = totEdge++; } void addQuery(int from, int to, int x) { Query[totQuery].to = to; Query[totQuery].num = x; Query[totQuery].next = headQuery[from]; headQuery[from] = totQuery++; } void init() { memset(headEdge, -1, sizeof(headEdge)); memset(headQuery, -1, sizeof(headQuery)); memset(father, -1, sizeof(father)); memset(vis, false, sizeof(vis)); memset(sum, 0, sizeof(sum)); memset(LCAnum, 0, sizeof(LCAnum)); totEdge = totQuery = 0; } int find_set(int x) { if(x == father[x]) return x; else return father[x] = find_set(father[x]); } void union_set(int x, int y) { x = find_set(x); y = find_set(y); if(x != y) father[y] = x; } void Tarjan(int u) { father[u] = u; for(int i = headEdge[u]; i != -1; i = edge[i].next) { int v = edge[i].to; if(father[v] != -1) continue; Tarjan(v); union_set(u, v); } for(int i = headQuery[u]; i != -1; i = Query[i].next) { int v = Query[i].to; if(father[v] == -1) continue; input[Query[i].num].lca = find_set(v); } } void DFS(int u, int pre) { vis[u] = 1; sum[u] = sum[pre] + LCAnum[u]; for(int i = headEdge[u]; i != -1; i = edge[i].next) { int v = edge[i].to; if(vis[v]) continue; DFS(v, u); } } int main() { init(); scanf("%d%d", &n, &m); for(int i = 0; i < n - 1; i++) { int a, b; scanf("%d%d", &a, &b); addEdge(a, b); addEdge(b, a); } for(int i = 0; i < m; i++) { int a, b; scanf("%d%d", &a, &b); input[i].u = a, input[i].v = b; addQuery(a, b, i); addQuery(b, a, i); } Tarjan(1); for(int i = 0; i < m; i++) LCAnum[input[i].lca]++; DFS(1, 0); LL ans = 0; for(int i = 0; i < m; i++) { ans += (sum[input[i].u] + sum[input[i].v] - 2 * sum[input[i].lca]); } for(int i = 1; i <= n; i++) { ans += (LL)LCAnum[i] * (LCAnum[i] - 1) / 2; } printf("%lld\n", ans); return 0; }
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原文地址:http://www.cnblogs.com/gaoxiang36999/p/4484389.html