Codeforces Round #359 (Div. 2)D. Kay and Snowflake
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题意:给定一棵以1为根的n个节点的树,然后m个询问,每次询问给定一个x。求在x为根的子树中的质心是谁。x的质心:在这颗子树中删掉它的质心,然后变成若干课小树,要求小树中的最大的size要<=x的size/2。
分析:我们直接预处理出每个点的质心,很容易想到x的质心一定在x和x的size最大的儿子的质心的路径上。为什么呢?画画图就知道了,不解释。然后我们只要dfs处理一遍就行啦。
求解时,从儿子节点的ans值向上找父节点,直到满足要求。
#include <set>#include <map>#include <stack>#include <queue>#include <deque>#include <cmath>#include <vector>#include <string>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define L(i) i<<1#define R(i) i<<1|1#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-9#define maxn 1000100#define MOD 1000000007struct Edge{ int to,next;} edge[maxn];int n,m,num[maxn],ans[maxn];int tot,head[maxn],pre[maxn];void init(){ tot = 0; memset(head,-1,sizeof(head));}void add_edge(int u,int v){ edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++;}void dfs(int u){ num[u] = 1; ans[u] = u; int mx = 0,tmp = u; if(head[u] == -1) return; for(int i = head[u]; i != -1; i = edge[i].next) { int v = edge[i].to; dfs(v); num[u] += num[v]; if(num[v] > mx) { mx = num[v]; tmp = v; } } ans[u] = ans[tmp]; while(ans[u] != u && num[u] - num[ans[u]] > num[u]/2) ans[u] = pre[ans[u]];}int main(){ int t; //scanf("%d",&t); while(scanf("%d%d",&n,&m) != EOF) { init(); for(int i = 2; i <= n; i++) { int x; scanf("%d",&x); pre[i] = x; add_edge(x,i); } dfs(1); while(m--) { int x; scanf("%d",&x); printf("%d\n",ans[x]); } } return 0;}
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