思路：树链剖分的模板
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
#define Del(a,b) memset(a,b,sizeof(a))
const int N = 10005;
int dep[N],siz[N],fa[N],id[N],son[N],val[N],top[N]; //top 最近的重链父节点
int num;
vector<int> v[N];
struct tree
{
int x,y,val;
void read(){
scanf("%d%d%d",&x,&y,&val);
}
};
tree e[N];
void dfs1(int u, int f, int d) {
dep[u] = d;
siz[u] = 1;
son[u] = 0;
fa[u] = f;
for (int i = 0; i < v[u].size(); i++) {
int ff = v[u][i];
if (ff == f) continue;
dfs1(ff, u, d + 1);
siz[u] += siz[ff];
if (siz[son[u]] < siz[ff])
son[u] = ff;
}
}
void dfs2(int u, int tp) {
top[u] = tp;
id[u] = ++num;
if (son[u]) dfs2(son[u], tp);
for (int i = 0; i < v[u].size(); i++) {
int ff = v[u][i];
if (ff == fa[u] || ff == son[u]) continue;
dfs2(ff, ff);
}
}
#define lson(x) ((x<<1))
#define rson(x) ((x<<1)+1)
struct Tree
{
int l,r,val;
};
Tree tree[4*N];
void pushup(int x) {
tree[x].val = max(tree[lson(x)].val, tree[rson(x)].val);
}
void build(int l,int r,int v)
{
tree[v].l=l;
tree[v].r=r;
if(l==r)
{
tree[v].val = val[l];
return ;
}
int mid=(l+r)>>1;
build(l,mid,v*2);
build(mid+1,r,v*2+1);
pushup(v);
}
void update(int o,int v,int val) //log(n)
{
if(tree[o].l==tree[o].r)
{
tree[o].val = val;
return ;
}
int mid = (tree[o].l+tree[o].r)/2;
if(v<=mid)
update(o*2,v,val);
else
update(o*2+1,v,val);
pushup(o);
}
int query(int x,int l, int r)
{
if (tree[x].l >= l && tree[x].r <= r) {
return tree[x].val;
}
int mid = (tree[x].l + tree[x].r) / 2;
int ans = 0;
if (l <= mid) ans = max(ans, query(lson(x),l,r));
if (r > mid) ans = max(ans, query(rson(x),l,r));
return ans;
}
int Yougth(int u, int v) {
int tp1 = top[u], tp2 = top[v];
int ans = 0;
while (tp1 != tp2) {
//printf("YES\n");
if (dep[tp1] < dep[tp2]) {
swap(tp1, tp2);
swap(u, v);
}
ans = max(query(1,id[tp1], id[u]), ans);
u = fa[tp1];
tp1 = top[u];
}
if (u == v) return ans;
if (dep[u] > dep[v]) swap(u, v);
ans = max(query(1,id[son[u]], id[v]), ans);
return ans;
}
void Clear(int n)
{
for(int i=1;i<=n;i++)
v[i].clear();
}
int main()
{
//freopen("Input.txt","r",stdin);
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
for(int i=1;i<n;i++)
{
e[i].read();
v[e[i].x].push_back(e[i].y);
v[e[i].y].push_back(e[i].x);
}
num = 0;
dfs1(1,0,1);
dfs2(1,1);
for (int i = 1; i < n; i++) {
if (dep[e[i].x] < dep[e[i].y]) swap(e[i].x, e[i].y);
val[id[e[i].x]] = e[i].val;
}
build(1,num,1);
char s[200];
while(~scanf("%s",&s) && s[0]!='D')
{
int x,y;
scanf("%d%d",&x,&y);
if(s[0]=='Q')
printf("%d\n",Yougth(x,y));
if (s[0] == 'C')
update(1,id[e[x].x],y);
}
Clear(n);
}
return 0;
}
Description
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:
- CHANGE i ti : change the cost of the i-th edge to ti
or - QUERY a b : ask for the maximum edge cost on the path from node a to node b
Input
The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.
For each test case:
- In the first line there is an integer N (N <= 10000),
- In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c<= 1000000),
- The next lines contain instructions "CHANGE i ti" or "QUERY a b",
- The end of each test case is signified by the string "DONE".
There is one blank line between successive tests.
Output
For each "QUERY" operation, write one integer representing its result.
Example
Input: 1 3 1 2 1 2 3 2 QUERY 1 2 CHANGE 1 3 QUERY 1 2 DONE Output: 1 3
Hint
Added by: | |
Date: | 2005-06-08 |
Time limit: | 0.851s |
Source limit: | 15000B |
Memory limit: | 1536MB |
Cluster: | |
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