【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
", ans[i]); 
    return 0; 
}