组队训练 K K

K - The Stream of Corning 2

这个题目不是很难，因为给你的这个S是单调递增的，所以就用优先队列+权值线段树就可以很快的解决了。

这个+读入挂可以优化，不过不用也没关系。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <bitset>
#include <string>
#include <algorithm>
#include <iostream>
#include <map>
#define inf 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3f
using namespace std;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
const int M = 1e6;
typedef long long ll;
int num[maxn * 4];
 
void build(int id, int l, int r) {
    num[id] = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(id << 1, l, mid);
    build(id << 1 | 1, mid + 1, r);
}
struct node {
    int l, r, val;
    node(int l=0,int r=0,int val=0):l(l),r(r),val(val){}
    bool operator<(const node &a)const {
        return a.r < r;
    }
};
priority_queue<node>que;
 
void update(int id, int l, int r, int pos, int f) {
    if (f) num[id]++;
    else num[id]--;
    if (l == r) return;
    int mid = (l + r) >> 1;
    if (pos <= mid) update(id << 1, l, mid, pos, f);
    else update(id << 1 | 1, mid + 1, r, pos, f);
}
 
int query(int id, int l, int r, int k) {
    if (l == r) return l;
    int mid = (l + r) >> 1;
    int val1 = num[id << 1];
    if (val1 >= k) return query(id << 1, l, mid, k);
    return query(id << 1 | 1, mid + 1, r, k - val1);
}
 
inline int read() {
    int X = 0; bool flag = 1; char ch = getchar();
    while (ch<'0' || ch>'9') { if (ch == '-') flag = 0; ch = getchar(); }
    while (ch >= '0'&&ch <= '9') { X = (X << 1) + (X << 3) + ch - '0'; ch = getchar(); }
    if (flag) return X;
    return ~(X - 1);
}
inline void write(int X) {
    if (X < 0) { putchar('-'); X = ~(X - 1); }
    int s[20], top = 0;
    while (X) { s[++top] = X % 10; X /= 10; }
    if (!top) s[++top] = 0;
    while (top) putchar(s[top--] + '0');
    putchar('
');
}
 
int main() {
    int t = read();
    for (int cas = 1; cas <= t; cas++) {
        printf("Case %d:
", cas);
        while (!que.empty()) que.pop();
        int n = read();
        memset(num, 0, sizeof(num));
        while (n--) {
            int opt;
            scanf("%d", &opt);
            if (opt == 1) {
                int l, val, r;
                l = read(), val = read(), r = read();
                update(1, 1, M, val, 1);
                que.push(node(l, r, val));
            }
            else {
                int s, k;
                s = read(), k = read();
                while (!que.empty()&&que.top().r < s) {
                    update(1, 1, M, que.top().val, 0);
                    que.pop();
                }
                if (k > que.size()) write(-1);
                else write(query(1, 1, M, k));
            }
        }
    }
    return 0;
}