题面

Sol

整体二分

比较简单，当练手题

每次树状数组统计

# include 
    
     # define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;const int _(3e5 + 5); IL ll Input(){    RG ll x = 0, z = 1; RG char c = getchar();    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);    return x * z;}int n, m, k, ql[_], qr[_], qv[_], ans[_];vector 
     
       son[_];struct Data{    int id;    ll v;} p[_], q1[_], q2[_];ll bit[_];IL void Modify(RG int x, RG ll v){    for(; x <= m; x += x & -x) bit[x] += v;}IL ll Query(RG int x){    RG ll ret = 0;    for(; x; x -= x & -x) ret += bit[x];    return ret;}IL void Solve(RG int l, RG int r, RG int L, RG int R){    if(L > R) return;    if(l == r){        for(RG int i = L; i <= R; ++i) ans[p[i].id] = l;        return;    }    RG int mid = (l + r) >> 1, t1 = 0, t2 = 0;    for(RG int i = l; i <= mid; ++i)        if(ql[i] <= qr[i])            Modify(ql[i], qv[i]), Modify(qr[i] + 1, -qv[i]);        else Modify(1, qv[i]), Modify(qr[i] + 1, -qv[i]), Modify(ql[i], qv[i]);    for(RG int i = L; i <= R; ++i){        RG ll sum = 0;        for(RG int j = 0, len = son[p[i].id].size(); j < len; ++j)            sum += Query(son[p[i].id][j]);        if(p[i].v <= sum) q1[++t1] = p[i];        else p[i].v -= sum, q2[++t2] = p[i];    }    for(RG int i = l; i <= mid; ++i)        if(ql[i] <= qr[i])            Modify(ql[i], -qv[i]), Modify(qr[i] + 1, qv[i]);        else Modify(1, -qv[i]), Modify(qr[i] + 1, qv[i]), Modify(ql[i], -qv[i]);    for(RG int i = L, j = 1; j <= t1; ++i, ++j) p[i] = q1[j];    for(RG int i = L + t1, j = 1; j <= t2; ++i, ++j) p[i] = q2[j];    Solve(l, mid, L, L + t1 - 1); Solve(mid + 1, r, L + t1, R);}int main(RG int argc, RG char* argv[]){    n = Input(); m = Input();    for(RG int i = 1; i <= m; ++i) son[Input()].push_back(i);    for(RG int i = 1; i <= n; ++i) p[i].id = i, p[i].v = Input();    k = Input();    for(RG int i = 1; i <= k; ++i)        ql[i] = Input(), qr[i] = Input(), qv[i] = Input();    ql[k + 1] = 1; qr[k + 1] = m; qv[k + 1] = 1e9;    Solve(1, k + 1, 1, n);    for(RG int i = 1; i <= n; ++i)        if(ans[i] == k + 1) puts("NIE");        else printf("%d\n", ans[i]);    return 0;}