Description
You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edges are numbered 1 through N − 1. Each edge is associated with a weight. Then you are to execute a series of instructions on the tree. The instructions can be one of the following forms:
CHANGE i v |
Change the weight of the ith edge to v |
NEGATE a b |
Negate the weight of every edge on the path from a to b |
QUERY a b |
Find the maximum weight of edges on the path from a to b |
Input
The input contains multiple test cases. The first line of input contains an integer t (t ≤ 20), the number of test cases. Then follow the test cases.
Each test case is preceded by an empty line. The first nonempty line of its contains N (N ≤ 10,000). The next N − 1 lines each contains three integers a, b and c, describing an edge connecting nodes a and b with weight c. The edges are numbered in the order they appear in the input. Below them are the instructions, each sticking to the specification above. A lines with the word “DONE” ends the test case.
Output
For each “QUERY” instruction, output the result on a separate line.
Sample Input
1 3 1 2 1 2 3 2 QUERY 1 2 CHANGE 1 3 QUERY 1 2 DONE
Sample Output
1 3
树链剖分模板题、好吧、我又逗比了，最开始初始化为-1，但是由于有负数的存在，所以一直不对、、受不了、总是改不了马虎的习惯
用线段树维护一个最大值和最小值。
#include <iostream> #include <algorithm> #include <cstdio> #include <queue> #include <cmath> #include <map> #include <iterator> #include <cstring> #include <string> using namespace std; #define INF 0x7fffffff #define ll long long #define N 100010 struct Edge2 { int a,b,c; }s[N<<1]; struct Edge { int to,next; }edge[N<<1]; int head[N],tot; int num[N]; int pos; int fa[N]; int son[N]; int p[N]; int fp[N]; int deep[N]; int size[N]; int top[N]; int n; void init() { tot=0; pos=1; memset(head,-1,sizeof(head)); memset(son,-1,sizeof(son)); } void add(int x,int y) { edge[tot].to=y; edge[tot].next=head[x]; head[x]=tot++; } void dfs1(int now,int pre,int d) { deep[now]=d; fa[now]=pre; size[now]=1; for(int i=head[now];i!=-1;i=edge[i].next) { int next=edge[i].to; if(next!=pre) { dfs1(next,now,d+1); size[now]+=size[next]; if(son[now]==-1 || size[next]>size[son[now]]) { son[now]=next; } } } } void dfs2(int now,int tp) { top[now]=tp; p[now]=pos++; fp[p[now]]=now; if(son[now]==-1) return; dfs2(son[now],tp); for(int i=head[now];i!=-1;i=edge[i].next) { int next=edge[i].to; if(next!=son[now]&&next!=fa[now]) { dfs2(next,next); } } } int mx[N<<2]; int mi[N<<2]; int col[N<<2]; void pushup(int rt) { mx[rt]=max(mx[rt<<1],mx[rt<<1|1]); mi[rt]=min(mi[rt<<1],mi[rt<<1|1]); } void pushdown(int rt) { if(col[rt]) { col[rt<<1]^=1; col[rt<<1|1]^=1; mx[rt<<1]=-mx[rt<<1]; mi[rt<<1]=-mi[rt<<1]; swap(mx[rt<<1],mi[rt<<1]); mx[rt<<1|1]=-mx[rt<<1|1]; mi[rt<<1|1]=-mi[rt<<1|1]; swap(mx[rt<<1|1],mi[rt<<1|1]); col[rt]=0; } } void build(int l,int r,int rt) { col[rt]=0; if(l==r) { mi[rt]=mx[rt]=num[fp[l]]; return; } int m=(l+r)>>1; build(l,m,rt<<1); build(m+1,r,rt<<1|1); pushup(rt); } void update(int l,int r,int rt,int pos,int val) { if(l==r) { mi[rt]=mx[rt]=val; return; } pushdown(rt); int m=(l+r)>>1; if(pos<=m) update(l,m,rt<<1,pos,val); else update(m+1,r,rt<<1|1,pos,val); pushup(rt); } void update2(int l,int r,int rt,int L,int R) { if(l==L && R==r) { mx[rt]=-mx[rt]; mi[rt]=-mi[rt]; swap(mx[rt],mi[rt]); col[rt]^=1; return; } pushdown(rt); int m=(l+r)>>1; if(R<=m) update2(l,m,rt<<1,L,R); else if(L>m) update2(m+1,r,rt<<1|1,L,R); else { update2(l,m,rt<<1,L,m); update2(m+1,r,rt<<1|1,m+1,R); } pushup(rt); } int query(int l,int r,int rt,int L,int R) { if(l==L && r==R) { return mx[rt]; } pushdown(rt); int m=(l+r)>>1,ans=-INF; if(L>m) return ans=max(ans,query(m+1,r,rt<<1|1,L,R)); else if(R<=m) ans=max(ans,query(l,m,rt<<1,L,R)); else return ans=max(ans,max(query(l,m,rt<<1,L,m),query(m+1,r,rt<<1|1,m+1,R))); return ans; } int convert(int pos) { int a=s[pos].a; int b=s[pos].b; if(deep[a]>deep[b]) return a; return b; } void pre_solve() { memset(num,-1,sizeof(num)); for(int i=1;i<n;i++) { num[convert(i)]=s[i].c; } } void change2(int x,int y) { int f1=top[x]; int f2=top[y]; while(f1!=f2) { if(deep[f1]<deep[f2]) { swap(x,y); swap(f1,f2); } update2(1,n,1,p[f1],p[x]); x=fa[f1]; f1=top[x]; } if(x==y) return; if(deep[x]>deep[y]) swap(x,y); update2(1,n,1,p[x]+1,p[y]); } int change(int x,int y) { int ans=-INF; int f1=top[x]; int f2=top[y]; while(f1!=f2) { if(deep[f1]<deep[f2]) { swap(x,y); swap(f1,f2); } ans=max(ans,query(1,n,1,p[f1],p[x])); x=fa[f1]; f1=top[x]; } if(x==y) return ans; if(deep[x]>deep[y]) swap(x,y); ans=max(ans,query(1,n,1,p[x]+1,p[y])); return ans; } int main() { int T; scanf("%d",&T); while(T--) { init(); scanf("%d",&n); for(int i=1;i<n;i++) { scanf("%d%d%d",&s[i].a,&s[i].b,&s[i].c); add(s[i].a,s[i].b); add(s[i].b,s[i].a); } dfs1(1,0,0); dfs2(1,1); pre_solve(); build(1,n,1); while(1) { int a,b; char op[10]; scanf("%s",op); if(op[0]=='D') break; scanf("%d%d",&a,&b); if(op[0]=='C') { a=convert(a); update(1,n,1,p[a],b); } else if(op[0]=='N') { change2(a,b); } else { printf("%d ",change(a,b)); } } } return 0; }