題面

bzoj
luogu

所有事件按時間排序
按值劃分下放

把每一個修改
改成一個刪除一個插入

對於一個查詢
直接查這個段區間有多少合法點
如果查詢值大於等於目標值 進入左區間

如果一個查詢無解
那麽它要求第k大無解
k > 路徑長 用lca維護即可

#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是查詢
}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！！先減再復制！！
        }
        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!\n"); 
            else printf("%d\n", ans[i]);
        } 
    //system("PAUSE");
    return 0;
}
 

[CTSC2008]網絡管理 [整體二分]