hiho1576 子树中的最小权值【dfs序】

来源：互联网发布：达科塔约翰逊知乎编辑：程序博客网时间：2024/05/16 05:34

题意：求以x为根的子树中的最小权值

思路：dfs序，把树转化成一个序列，类似LCA中的转化，记录1棵子树进去的时间戳和出去的时间戳。线段树求区间最小值

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<stdlib.h>#include<math.h>#include<vector>#include<list>#include<map>#include<set>#include<stack>#include<queue>#include<algorithm>#include<numeric>#include<functional>using namespace std;#define lson(x) 2*x#define rson(x) 2*x+1typedef long long ll;typedef pair<int,int> pii;const int maxn = 100005;int in[maxn],out[maxn];struct Edge{int to,next;}edge[maxn];int head[maxn],tot,t;int a[maxn],num[maxn];void init(){memset(head,-1,sizeof head);tot = 0;t = 0;}void add(int x,int y){edge[tot].to = y;edge[tot].next = head[x];head[x] = tot++;}void dfs(int x){in[x] = ++t;num[t] = a[x];for(int i = head[x]; i != -1; i = edge[i].next){int y = edge[i].to;dfs(y);}out[x] = t;}struct data{int l,r,num,laz;}node[maxn*4];void pushup(int cnt){node[cnt].num = min(node[lson(cnt)].num,node[rson(cnt)].num);}void build(int x,int y, int cnt){node[cnt].l = x;node[cnt].r = y;if(x == y){node[cnt].num = num[x];return;}int mid = (x+y) / 2;build(x,mid,lson(cnt));build(mid+1,y,rson(cnt));pushup(cnt);}void up(int x,int y,int cnt,int val){if(x == node[cnt].l && y == node[cnt].r){node[cnt].num = val;return;}int mid = (node[cnt].l + node[cnt].r) / 2;if(y <= mid)up(x,y,lson(cnt),val);else if(x >= mid+1)up(x,y,rson(cnt),val);else{up(x,mid,lson(cnt),val);up(mid+1,y,rson(cnt),val);}pushup(cnt);}int fid(int x,int y,int cnt){if(x == node[cnt].l && y == node[cnt].r){return node[cnt].num;}int mid = (node[cnt].l + node[cnt].r) / 2;if(y <= mid)return fid(x,y,lson(cnt));else if(x >= mid+1)return fid(x,y,rson(cnt));elsereturn min( fid(x,mid,lson(cnt)) , fid(mid+1,y,rson(cnt)));}int main(void){int n;while(scanf("%d",&n)!=EOF){init();for(int i = 1; i <= n; i++)scanf("%d",&a[i]);for(int i = 2; i <= n; i++){int k; scanf("%d",&k);add(k,i);}dfs(1);build(1,n,1);int q;scanf("%d",&q);while(q--){int op,x,u;scanf("%d",&op);if(op == 1){scanf("%d%d",&x,&u);up(in[x],in[x],1,u);}else{scanf("%d",&x);printf("%d\n",fid(in[x],out[x],1));}}}return 0;}

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