【LG3527】[POI2011]MET-Meteors

【LG3527】[POI2011]MET-Meteors

题面

洛谷

题解

整体二分。

每次二分\(mid\)，如果到时间\(mid\)以收集过\(P_i\)就存入子序列\(L\)，否则存入子序列\(R\)

修改可以树状数组区间修改单点查询做

每个王国的掉落地点用\(vector\)存一下即可

看起来复杂度是平方的实则为线性的

总复杂度\(O(nlog^2)\)

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
namespace IO {
    const int BUFSIZE = 1 << 20;
    char ibuf[BUFSIZE], *is = ibuf, *it = ibuf;
    inline char gc() {
        if (is == it) it = (is = ibuf) + fread(ibuf, 1, BUFSIZE, stdin);
		return *is++;
    }
}
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (ch != '-' && (ch > '9' || ch < '0')) ch = IO::gc();
    if (ch == '-') w = -1 , ch = IO::gc();
    while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = IO::gc();
    return w * data;
}
typedef long long ll;
const int MAX_N = 300005;
struct Node { ll p; int id; } q[MAX_N], lq[MAX_N], rq[MAX_N];
struct Option { int l, r, v; } p[MAX_N];
vector<int> G[MAX_N];
int N, M, K, ans[MAX_N];
ll c[MAX_N];
inline int lb(int x) { return x & -x; }
void add(int x, int v) { while (x <= M) c[x] += v, x += lb(x); }
ll sum(int x) { ll res = 0; while (x > 0) res += c[x], x -= lb(x); return res; }
void modify(int x, int w) {
    int l = p[x].l, r = p[x].r, v = p[x].v;
    if (l <= r) add(l, w * v), add(r + 1, -w * v);
    else add(1, v * w), add(r + 1, -v * w), add(l, v * w);
}
void Div(int lval, int rval, int st, int ed) {
	if (st > ed) return ;
	if (lval == rval) { for (int i = st; i <= ed; i++) ans[q[i].id] = lval; return ; }
	int mid = (lval + rval) >> 1;
	int lt = 0, rt = 0; ll res = 0;
	for (int i = lval; i <= mid; i++) modify(i, 1);
	for (int i = st; i <= ed; i++) {
	    res = 0; vector<int> :: iterator ite; int x = q[i].id;
	    for (ite = G[x].begin(); ite != G[x].end(); ++ite) { res += sum(*ite); if (res >= q[i].p) break; }
        if (res >= q[i].p) lq[++lt] = q[i];
	    else q[i].p -= res, rq[++rt] = q[i];
    }
    for (int i = lval; i <= mid; i++) modify(i, -1);
    for (int i = 1; i <= lt; i++) q[st + i - 1] = lq[i];
    for (int i = 1; i <= rt; i++) q[st + lt + i - 1] = rq[i];
    Div(lval, mid, st, st + lt - 1);
    Div(mid + 1, rval, st + lt, ed);
}
int main () {
    N = gi(), M = gi();
    for (int i = 1; i <= M; i++) G[gi()].push_back(i);
    for (int i = 1; i <= N; i++) q[i] = (Node){gi(), i};
    K = gi();
    for (int i = 1; i <= K; i++) p[i] = (Option){gi(), gi(), gi()};
    ++K; p[K] = (Option){1, M, 1e9};
    Div(1, K, 1, N);
    for (int i = 1; i <= N; i++) ans[i] == K ? puts("NIE") : printf("%d\n", ans[i]);
    return 0;
}