牛客练习赛48 E 小w的矩阵前k大元素

E

思路：

优先队列，将迭代器变量作为结构体的变量。

迭代器走的时候只能像一个方向走，另外一个方向只有最开始才走。如下图所示：

如果两个方向同时走，同一个值会被遍历多次，像上图那样就能保证每个位置都走到且只走一次。

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

const int N = 1e5 + ;
int n, m, q, k;
int a[N], b[N];
char op[];
multiset<int> sa, sb;
struct Node {
    int x;
    multiset<int>::iterator pa, pb;
    bool operator < (const Node & rhs) const {
        return x > rhs.x;
    }
};
int main() {
    scanf("%d %d %d", &n, &m, &q);
    for (int i = ; i <= n; ++i) scanf("%d", &a[i]);
    for (int i = ; i <= m; ++i) scanf("%d", &b[i]);
    int x = , y = ;
    sa.insert(a[]), sb.insert(b[]);
    while(q--) {
        scanf("%s %d", op, &k);
        if(op[] == 'R') {
            int ny = min(m, y+k);
            for (int i = y+; i <= ny; ++i) sb.insert(b[i]);
            y = ny;
        }
        else if(op[] == 'D') {
            int nx = min(n, x+k);
            for (int i = x+; i <= nx; ++i) sa.insert(a[i]);
            x = nx;
        }
        else {
            priority_queue<Node> q;
            q.push({*sa.begin()+*sb.begin(), sa.begin(), sb.begin()});
            for (int i = ; i <= k; ++i) {
                Node t = q.top();
                q.pop();
                printf("%d%c", t.x, " \n"[i==k]);
                if(t.pa == sa.begin()) {
                    auto tp = t.pb;
                    ++tp;
                    if(tp != sb.end()) q.push({*t.pa+*tp, t.pa, tp});
                }
                auto ta = t.pa;
                ++ta;
                if(ta != sa.end()) q.push({*ta+*t.pb, ta, t.pb});
            }
        }
    }
    return ;
}