SP375 Query on a tree【树链剖分】

题意：给定一棵树，求一条路径上的边权最大值

思路：树链剖分，把一颗树，转化成一个区间，就可以用线段树，快速解决一些问题了

至此用这个模板终于会处理点权问题了，LCA那个dfs序转成链，我不会算点权的

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<stdlib.h>#include<math.h>#include<vector>#include<list>#include<map>#include<stack>#include<queue>#include<algorithm>#include<numeric>#include<functional>using namespace std;typedef long long ll;typedef pair<int,int> pii;#define lson(x) 2*x#define rson(x) 2*x+1const int maxn = 1e4+5;struct Edge{int to,next,cost;}edge[maxn*4];int e1[maxn],e2[maxn],cost[maxn];int head[maxn],tot,cnt;int root,size[maxn],dep[maxn],fa[maxn],son[maxn];int top[maxn],id[maxn],val[maxn];/*size -> 以x为根节点子树的大小   dep -> 深度 fa   -> 父节点                  son -> 重儿子 top  -> x所在重链的顶端节点     id -> 这个点在线段树中的位置 */struct data{int l,r,num;}node[maxn*4];void init(){memset(head,-1,sizeof head);tot = 0;cnt = 0;}void add(int x,int y,int val){edge[tot].to = y;edge[tot].cost = val;edge[tot].next = head[x];head[x] = tot++;}int dfs1(int x,int pre,int d){size[x] = 1;fa[x] = pre;dep[x] = d;son[x] = 0;for(int i = head[x]; i != -1; i = edge[i].next){int y = edge[i].to;if(y != pre){size[x] += dfs1(y,x,d+1);if(size[y] > size[son[x]])son[x] = y;}}return size[x];}void dfs2(int x,int pre){id[x] = ++cnt;top[x] = pre;if(son[x]) dfs2(son[x],pre); //重链 for(int i = head[x]; i != -1; i = edge[i].next){int y = edge[i].to;if(y != fa[x] && y != son[x])dfs2(y,y); //轻链}}void pushup(int cnt){node[cnt].num = max(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 = val[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 fa = 2*cnt;if(x <= node[fa].r){if(y <= node[fa].r)up(x,y,fa,val);elseup(x,node[fa].r,fa,val);}fa++;if(y >= node[fa].l){if(x >= node[fa].l)up(x,y,fa,val);elseup(node[fa].l,y,fa,val);}pushup(cnt);}int fid(int x,int y,int cnt){int res = 0;if(x == node[cnt].l && y == node[cnt].r){return node[cnt].num;}int fa = 2*cnt;if(x <= node[fa].r){if(y <= node[fa].r)res = max(res,fid(x,y,fa));elseres = max(res,fid(x,node[fa].r,fa));}fa++;if(y >= node[fa].l){if(x >= node[fa].l)res = max(res,fid(x,y,fa));elseres = max(res,fid(node[fa].l,y,fa));}return res;}int qurey(int u,int v){int ans = 0;while(top[u] != top[v]){if(dep[top[u]] < dep[top[v]])swap(u,v);ans = max(ans,fid(id[top[u]],id[u],1));u = top[u];u = fa[u];}if(dep[u] > dep[v])swap(u,v);ans = max(ans,fid(id[son[u]],id[v],1)); return ans;//存的都是点的父边，最上面的点用son[u] }int main(){int T,n,kase = 1;char ch[50];scanf("%d",&T);while(T--){if(kase++ != 1)putchar('\n');scanf("%d",&n);init();for(int i = 1; i < n; i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);e1[i] = a;e2[i] = b;cost[i] = c;add(a,b,c);add(b,a,c);}root = (n+1) / 2;dfs1(root,root,1);dfs2(root,root);for(int i = 1; i < n; i++){if(dep[ e1[i] ] < dep[ e2[i] ])swap(e1[i],e2[i]);val[ id[e1[i]] ] = cost[i];}build(1,cnt,1);while(scanf("%s",ch)!=EOF && ch[0]!='D'){int a,b;scanf("%d%d",&a,&b);if(ch[0] == 'Q'){int ans = qurey(a,b);printf("%d\n",ans);}elseup(id[e1[a]],id[e1[a]],1,b);}}}