【LG3527】[POI2011]MET-Meteors
【LG3527】[POI2011]MET-Meteors
题面
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;
}
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