「POI2011」Meteors

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「POI2011」Meteors

传送门
整体二分，树状数组实现区间修改单点查询，然后注意修改是在环上的。
参考代码：

#include <cstdio>
#include <vector> 
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
using namespace std;

template < class T > inline void read(T& s) {
    s = 0; rg int f = 0; rg char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}

const int _ = 9e5 + 5;

int n, m, k, p[_], tr[_], ans[_]; vector < int > pos[_];
int num; struct node { int opt, l, r, x, id;  } t[_], tt1[_], tt2[_];

inline void update(int x, int v) { for (rg int i = x; i <= m; i += i & -i) tr[i] += v; }

inline int query(int x) { int res = 0; for (rg int i = x; i >= 1; i -= i & -i) res += tr[i]; return res; }

inline void solve(int ql, int qr, int l, int r) {
    if (ql > qr || l > r) return ;
    if (l == r) { for (rg int i = ql; i <= qr; ++i) if (t[i].opt == 0) ans[t[i].id] = l; return ; }
    int mid = (l + r) >> 1, p1 = 0, p2 = 0;
    for (rg int i = ql; i <= qr; ++i) {
        if (t[i].opt != 0) {
            if (t[i].id <= mid) {
                if (t[i].opt == 1) update(t[i].l, t[i].x), update(t[i].r + 1, -t[i].x);
                else update(1, t[i].x), update(t[i].r + 1, -t[i].x), update(t[i].l, t[i].x);
                tt1[++p1] = t[i];
            } else tt2[++p2] = t[i];
        } else {
            int cnt = 0;
            for (rg int j = 0; j < pos[t[i].id].size(); ++j) {
                cnt += query(pos[t[i].id][j]); if (cnt >= t[i].x) break ;
            }
            if (cnt >= t[i].x) tt1[++p1] = t[i]; else t[i].x -= cnt, tt2[++p2] = t[i];
        }
    }
    for (rg int i = 1; i <= p1; ++i)
        if (tt1[i].opt != 0) {
            if (tt1[i].opt == 1) update(tt1[i].l, -tt1[i].x), update(tt1[i].r + 1, tt1[i].x);
            else update(1, -tt1[i].x), update(tt1[i].r + 1, tt1[i].x), update(tt1[i].l, -tt1[i].x);
        }
    for (rg int i = 1; i <= p1; ++i) t[ql + i - 1] = tt1[i];
    for (rg int i = 1; i <= p2; ++i) t[ql + p1 + i - 1] = tt2[i];
    solve(ql, ql + p1 - 1, l, mid), solve(ql + p1, qr, mid + 1, r);
}

int main() {
    read(n), read(m);
    for (rg int x, i = 1; i <= m; ++i) read(x), pos[x].push_back(i);
    for (rg int i = 1; i <= n; ++i) read(p[i]);
    read(k);
    for (rg int opt, l, r, x, i = 1; i <= k; ++i) {
        read(l), read(r), read(x);
        if (l <= r) t[++num] = (node) { 1, l, r, x, i }; else t[++num] = (node) { 2, l, r, x, i };
    }
    for (rg int i = 1; i <= n; ++i) t[++num] = (node) { 0, 0, 0, p[i], i };
    solve(1, num, 1, k + 1);
    for (rg int i = 1; i <= n; ++i) if (ans[i] == k + 1) puts("NIE"); else printf("%d\n", ans[i]);
    return 0;
}

「POI2011」Meteors

标签：har   put   代码   clu   else   ons   http   ++i   tps   

原文地址：https://www.cnblogs.com/zsbzsb/p/12231710.html