[luoguP2713] 罗马游戏

题意

原题链接
维护一个数据结构,要求支持合并集合或删除集合最小值并输出。

sol

双倍经验,同 [luoguP3377] 左偏树/可并堆

代码

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

const int N = 1000005;

struct Node{
    int l, r;
    int val;
    int dist;
} tr[N];
int n, m;
bool st[N];
int fa[N];

int find(int x){
    if (fa[x] == x) return x;
    return fa[x] = find(fa[x]);
}

int merge(int x, int y){
    if (!x || !y) return x | y;
    if (tr[x].val > tr[y].val) swap(x, y);
    tr[x].r = merge(tr[x].r, y);
    if (tr[tr[x].l].dist < tr[tr[x].r].dist) swap(tr[x].l, tr[x].r);
    tr[x].dist = tr[tr[x].r].dist + 1;
    return x;
}

int main(){
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++ ) scanf("%d", &tr[i].val), fa[i] = i;
    scanf("%d", &m);
    while (m -- ){
        char op[2];
        int x, y;
        scanf("%s", op);
        if (*op == 'M') {
            scanf("%d%d", &x, &y);
            int fx = find(x), fy = find(y);
            if (st[x] || st[y] || fx == fy) continue;
            fa[fx] = fa[fy] = merge(fx, fy);
        }
        else {
            scanf("%d", &x);
            int fx = find(x);
            if (st[x]) puts("0");
            else {
                printf("%d\n", tr[fx].val);
                st[fx] = true;
                fa[tr[fx].l] = fa[tr[fx].r] = fa[fx] = merge(tr[fx].l, tr[fx].r);
            }
        }
    }
}
posted @ 2024-11-06 21:42  是一只小蒟蒻呀  阅读(1)  评论(0编辑  收藏  举报