题目链接
给一棵树， m个操作， 每种操作将节点u, v之间的节点全都加上一个值c， 1<=c<=1e5. 问你最后每种节点上面， 哪一种值c出现的次数最多。
我们将树转换成链， 然后对于每种操作， 可以写成add[u].push_back(c), sub[v+1].push_back(c）。
代表u到v之间所有值都加上了c。
最后统计每一种值c的出现次数的时候用线段树统计就可以了。
#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <complex>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef complex <double> cmx;
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 100005;
int head[maxn], son[maxn], sz[maxn], deep[maxn], top[maxn], w[maxn], f[maxn], fp[maxn], cnt, num;
int maxx[maxn<<2], ans[maxn];
vector <int> inc[maxn], des[maxn];
struct node
{
int to, nextt;
}e[maxn*2];
void init() {
mem1(head);
num = cnt = 0;
}
void add(int u, int v) {
e[num].to = v, e[num].nextt = head[u], head[u] = num++;
}
void dfs1(int u, int fa) {
sz[u] = 1;
deep[u] = deep[fa]+1;
son[u] = -1;
f[u] = fa;
for(int i = head[u]; ~i; i = e[i].nextt) {
int v = e[i].to;
if(v == fa)
continue;
dfs1(v, u);
sz[u] += sz[v];
if(son[u]==-1||sz[v]>sz[son[u]])
son[u] = v;
}
}
void dfs2(int u, int tp) {
w[u] = ++cnt, top[u] = tp;
fp[cnt] = u;
if(~son[u])
dfs2(son[u], tp);
for(int i = head[u]; ~i; i = e[i].nextt) {
int v = e[i].to;
if(v == f[u]||v == son[u])
continue;
dfs2(v, v);
}
}
void change(int u, int v, int c) {
int t1 = top[u], t2 = top[v];
while(t1 != t2) {
if(deep[t1] < deep[t2]) {
swap(t1, t2);
swap(u, v);
}
inc[w[t1]].pb(c);
des[w[u]+1].pb(c);
u = f[t1];
t1 = top[u];
}
if(deep[u]>deep[v])
swap(u, v);
inc[w[u]].pb(c);
des[w[v]+1].pb(c);
}
void update(int p, int val, int l, int r, int rt) {
if(l == r) {
maxx[rt] += val;
return ;
}
int m = l+r>>1;
if(p <= m)
update(p, val, lson);
else
update(p, val, rson);
maxx[rt] = max(maxx[rt<<1], maxx[rt<<1|1]);
}
int query(int l, int r, int rt) {
if(l == r) {
return l;
}
int m = l+r>>1;
if(maxx[rt<<1] >= maxx[rt<<1|1])
return query(lson);
return query(rson);
}
int main()
{
int n, m, u, v, w;
while(cin>>n>>m) {
if(n+m==0)
break;
init();
for(int i = 0; i < n-1; i ++) {
scanf("%d%d", &u, &v);
add(u, v);
add(v, u);
}
dfs1(1, 0);
dfs2(1, 1);
for(int i = 1; i <= n; i++) {
inc[i].clear();
des[i].clear();
}
for(int i = 0; i < m; i++) {
scanf("%d%d%d", &u, &v, &w);
change(u, v, w);
}
mem(maxx);
for(int i = 1; i <= n; i++) {
for(int j = 0; j < inc[i].size(); j++) {
update(inc[i][j] , 1, 1, 1e5, 1);
}
for(int j = 0; j < des[i].size(); j++) {
update(des[i][j], -1, 1, 1e5, 1);
}
int tmp = maxx[1]?query(1, 1e5, 1):0;
ans[fp[i]] = tmp;
}
for(int i = 1; i <= n; i++) {
printf("%d
", ans[i]);
}
}
return 0;
}