# BZOJ1455罗马游戏

（搞板子搞到绝望x

//罗马游戏
#include <bits/stdc++.h>
using namespace std;
const int N = 1000050;
int val[N], l[N], r[N], d[N], fa[N];
int x, y, n, m;
char ord;
// 合并以 x, y为根的堆，返回合并后的堆顶
int merge(int x, int y)
{
if(!x||!y) return x|y;
// 键值小的作为堆顶
if (val[x] > val[y])    swap(x, y);
r[x] = merge(r[x], y);
// 维护左偏性质
if (d[r[x]] > d[l[x]])
swap(l[x], r[x]);
// 重新计算距离
d[x] = d[r[x]] + 1;
return x;
}

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

int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d", &val[i]);
fa[i] = i;
}

scanf("%d", &m);
for (int i = 1; i <= m; i++) {
cin >> ord;
if (ord == 'M') {
scanf("%d%d", &x, &y);
if (val[x] < 0 || val[y] < 0) continue;
x = find(x), y = find(y);
if (x != y) {
int t = merge(x, y);
fa[x] = fa[y] = t;
}
}
else {
scanf("%d", &x);
if (val[x] < 0) {
puts("0");
continue;
}
x = find(x);
printf("%d\n", val[x]);
val[x] = -1;
fa[x] = merge(l[x], r[x]);
fa[fa[x]] = fa[x];
}
}
return 0;
}

posted @ 2019-10-03 20:59  QUEKI嶺冬  阅读(149)  评论(0编辑  收藏  举报
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