题面
所有事件按时间排序
按值划分下放
把每一个修改
改成一个删除一个插入
对于一个查询
直接查这个段区间有多少合法点
如果查询值大于等于目标值 进入左区间
如果一个查询无解
那么它要求第k大无解
k > 路径长 用lca维护即可
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
#define Sqr(x) ((x)*(x))
using namespace std;
const int N = 8e4 + 5;
const int inf = 1e8 + 5;
struct Edge{
int v, next;
}edge[N << 1];
int head[N], esize;
inline void addedge(int x, int y){
edge[++esize] = (Edge){y, head[x]};
head[x] = esize;
}
struct Node{
int t, x, y, w, d;
bool type; //0是插入 1是查询
}node[N << 2], p1[N << 2], p2[N << 2];
int n, m, nsize, a[N], ans[N];
int tim, dfn[N], top[N], son[N], fa[N], size[N], rf[N], dep[N];
bool flag[N];
struct BIT{
int w[N];
void ins(int x, int d){while(x <= n){w[x] += d; x += x & -x;}}
int qry(int x){int res = 0; while(x){res += w[x]; x -= x & -x;} return res;}
}bit;
inline void addnode(int x1, int x2, int x3, int x4, int x5, bool x6){
++nsize, node[nsize].t = x1, node[nsize].x = x2, node[nsize].y = x3,
node[nsize].w = x4, node[nsize].d = x5, node[nsize].type = x6;
}
void dfs1(int x, int ff){
dep[x] = dep[ff] + 1, size[x] = 1, fa[x] = ff;
for(int i = head[x], vv; ~i; i = edge[i].next){
vv = edge[i].v; if(vv == ff) continue;
dfs1(vv, x);
size[x] += size[vv];
if(size[son[x]] < size[vv]) son[x] = vv;
}
}
void dfs2(int x, int tp){
top[x] = tp, dfn[x] = ++tim, rf[tim] = x;
if(son[x]) dfs2(son[x], tp); else return;
for(int i = head[x], vv; ~i; i = edge[i].next){
vv = edge[i].v; if(vv == fa[x] || vv == son[x]) continue;
dfs2(vv, vv);
}
}
int LCA(int x, int y){
while(top[x] != top[y]){
if(dep[top[x]] < dep[top[y]]) swap(x, y);
x = fa[top[x]];
}
return dep[x] < dep[y] ? x : y;
}
int qry(int x, int y){
int res = 0;
while(top[x] != top[y]){
if(dep[top[x]] < dep[top[y]]) swap(x, y);
res += bit.qry(dfn[x]) - bit.qry(dfn[top[x]] - 1);
x = fa[top[x]];
}
if(dep[x] < dep[y]) swap(x, y); res += bit.qry(dfn[x]) - bit.qry(dfn[y] - 1);
return res;
}
void erfn(int L, int R, int l, int r){
if(L > R) return;
if(l == r){
for(int i = L; i <= R; ++i)
if(node[i].type && ~ans[node[i].t]) ans[node[i].t] = l;
return ;
}
int mid = l + ((r - l) >> 1);
int t1 = 0, t2 = 0;
for(int i = L; i <= R; ++i){
if(node[i].type) {
int cnt = qry(node[i].x, node[i].y);
if(cnt >= node[i].w) p2[++t2] = node[i];
else node[i].w -= cnt, p1[++t1] = node[i];//第k大！！先减再复制！！
}
else {
if(node[i].w <= mid) p1[++t1] = node[i];
else p2[++t2] = node[i], bit.ins(dfn[node[i].x], node[i].d);
}
}
for(int i = L; i <= R; ++i)
if(!node[i].type && node[i].w > mid)
bit.ins(dfn[node[i].x], -node[i].d);
for(int i = 1; i <= t1; ++i) node[L + i - 1] = p1[i];
for(int i = 1; i <= t2; ++i) node[L + t1 + i - 1] = p2[i];
erfn(L, L + t1 - 1, l, mid); erfn(L + t1, R, mid + 1, r);
}
int main() {
memset(head, -1, sizeof(head));
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; ++i){
scanf("%d", &a[i]);
addnode(0, i, 0, a[i], 1, 0);
}
for(int i = 1, x, y; i < n; ++i){
scanf("%d%d", &x, &y);
addedge(x, y); addedge(y, x);
} dfs1(1, 0); dfs2(1, 1);
for(int i = 1, x, y, z; i <= m; ++i){
scanf("%d%d%d", &x, &y, &z); flag[i] = (bool)x;
if(x){
int len = dep[y] + dep[z] - 2 * dep[LCA(y, z)] + 1;
if(x > len) ans[i] = -1;
else addnode(i, y, z, x, 0, 1);
}
else addnode(i, y, 0, a[y], -1, 0), addnode(i, y, 0, z, 1, 0), a[y] = z, ans[i] = -1;
}
erfn(1, nsize, 0, inf);
for(int i = 1; i <= m; ++i)
if(flag[i]){
if(ans[i] == -1) printf("invalid request!
");
else printf("%d
", ans[i]);
}
//system("PAUSE");
return 0;
}