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You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1.
We will ask you to perfrom some instructions of the following form:
The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.
For each test case:
There is one blank line between successive tests.
For each "QUERY" operation, write one integer representing its result.
Input:
1
3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE
Output:
1
3
模板题
2种操作：
1.把第i条边的边权更改为v
2.查询路径u，v中最大的边权
注意：是边权，所以查询时候u，v的lca不能包括进来
因为lca所表示的边权并没有在u，v的路径中
树链剖分+线段树(单点更新，区间查询)
线段树连lazy都不用
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define LL long long #define debug printf("eeeeeeeeeeeeeeeeeeeee") #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 using namespace std; const int maxn=1e4+10; const int inf=0x3f3f3f3f; int dep[maxn]; int son[maxn]; int siz[maxn]; int top[maxn]; int fa[maxn]; int chg[maxn]; int rev[maxn]; int cost[maxn]; int e[maxn][3]; struct Edge { int to,next; }; Edge edge[maxn<<1]; int head[maxn]; int tot; int ma[maxn<<2]; void init() { memset(head,-1,sizeof head); tot=0; } void addedge(int u,int v) { edge[tot].to=v; edge[tot].next=head[u]; head[u]=tot++; } void solve(int ); int main() { int test; scanf("%d",&test); while(test--){ int n; init(); scanf("%d",&n); for(int i=1;i<n;i++){ scanf("%d %d %d",&e[i][0],&e[i][1],&e[i][2]); addedge(e[i][0],e[i][1]); addedge(e[i][1],e[i][0]); } solve(n); } return 0; } void dfs0(int u,int pre) { fa[u]=pre; siz[u]=1; dep[u]=dep[pre]+1; for(int i=head[u];~i;i=edge[i].next){ int v=edge[i].to; if(v==pre) continue; dfs0(v,u); siz[u]+=siz[v]; if(son[u]==-1 || siz[v]>siz[son[u]]) son[u]=v; } } void dfs1(int u,int tp) { top[u]=tp; chg[u]=++tot; rev[tot]=u; if(son[u]==-1) return ; dfs1(son[u],tp); for(int i=head[u];~i;i=edge[i].next){ int v=edge[i].to; if(v==fa[u] || v==son[u]) continue; dfs1(v,v); } } void pushup(int rt) { ma[rt]=max(ma[rt<<1],ma[rt<<1|1]); } void build(int l,int r,int rt) { if(l==r){ ma[rt]=cost[rev[l]]; return ; } int m=(l+r)>>1; build(lson); build(rson); pushup(rt); } void update(int p,int add,int l,int r,int rt) { if(l==r){ ma[rt]=add; return ; } int m=(l+r)>>1; if(p<=m) update(p,add,lson); else update(p,add,rson); pushup(rt); } int query(int L,int R,int l,int r,int rt) { if(L<=l && R>=r){ return ma[rt]; } int m=(l+r)>>1; int ret=-inf; if(L<=m) ret=max(ret,query(L,R,lson)); if(R>m) ret=max(ret,query(L,R,rson)); return ret; } int query_path(int u,int v) { int ret=-inf; while(top[u]!=top[v]){ if(dep[top[u]]<dep[top[v]]) swap(u,v); ret=max(ret,query(chg[top[u]],chg[u],1,tot,1)); u=fa[top[u]]; } if(dep[u]<dep[v]) swap(u,v); ret=max(ret,query(chg[v]+1,chg[u],1,tot,1)); return ret; } void solve(int n) { memset(dep,0,sizeof dep); memset(son,-1,sizeof son); memset(cost,0,sizeof cost); dfs0(1,1); tot=0; dfs1(1,1); for(int i=1;i<n;i++){ if(dep[e[i][0]]>dep[e[i][1]]) swap(e[i][0],e[i][1]); cost[e[i][1]]=e[i][2]; } build(1,tot,1); char str[15]; while(scanf("%s",str)){ if(str[0]==‘D‘) break; int u,v; scanf("%d %d",&u,&v); if(str[0]==‘Q‘){ printf("%d\n",query_path(u,v)); } else{ update(chg[e[u][1]],v,1,tot,1); } } return ; }
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原文地址：http://www.cnblogs.com/-maybe/p/4883363.html
