思路&心路
一眼认定***提
写的比较慢,写了1小时吧
开心的交上去
卧槽,只有20?
不服不服,拿着题解的代码去对拍
Emma,<100没问题
100000数据错了,还只是错了一个数据
debug 啊debug啊
以为是数据是0的锅
终于早出了小样例
卧槽,这std不对啊,md
又换了个std
绝望的乱改一通
最后发现是tm vis数组开小了
其实第一遍就能过的
╮(╯▽╰)╭
代码
#include <iostream>
#include <cstdio>
#define ll long long
#define ls rt << 1
#define rs rt << 1 | 1
using namespace std;
const int maxn = 1e5 + 7;
const int maxm = 4e5 + 7;
const int inf = 0x3f3f3f3f;
bool vis[maxn*10];
int js,gs;
struct node {
int l, r, ma, mi;
ll meili, jiazhi;
} e[maxm];
int read() {
int x = 0, f = 1; char s = getchar();
for (; s > '9' || s < '0'; s = getchar()) if (s == '-') f = -1;
for (; s >= '0' && s <= '9'; s = getchar()) x = (x << 3) + (x << 1) + s - '0';
return x * f;
}
void pushup(int rt) {
e[rt].ma = max(e[ls].ma, e[rs].ma);
e[rt].mi = min(e[ls].mi, e[rs].mi);
e[rt].meili = e[ls].meili + e[rs].meili;
e[rt].jiazhi= e[ls].jiazhi+ e[rs].jiazhi;
}
void build(int l, int r, int rt) {
e[rt].l = l,e[rt].r = r,e[rt].mi = inf;
if(l == r) return;
int mid=(l + r) >> 1;
build(l,mid,ls);
build(mid+1,r,rs);
pushup(rt);
}
void update(int L, int w, int c, int rt) {
if(e[rt].l == e[rt].r) {
e[rt].meili = w;
e[rt].jiazhi = c;
e[rt].ma = e[rt].mi = c;
return;
}
int mid=(e[rt].l + e[rt].r) >> 1;
if(L <= mid) update(L, w, c, ls);
else update(L, w, c, rs);
pushup(rt);
}
void find_min(int rt) {
if(e[rt].l == e[rt].r) {
vis[e[rt].jiazhi] = 0;
e[rt].meili = 0;
e[rt].jiazhi = 0;
e[rt].ma = 0;
e[rt].mi = inf;
return;
}
if(e[ls].mi < e[rs].mi) find_min(ls);
else find_min(rs);
pushup(rt);
}
void find_max(int rt) {
if(e[rt].l == e[rt].r) {
vis[e[rt].jiazhi] = 0;
e[rt].meili = 0;
e[rt].jiazhi = 0;
e[rt].ma = 0;
e[rt].mi = inf;
return;
}
if(e[ls].ma > e[rs].ma) find_max(ls);
else find_max(rs);
pushup(rt);
}
int main() {
build(1, 1e5, 1);
while (233) {
int tmp = read();
if (tmp == 1) {
int a = read(),b = read();
if(!vis[b]) {
vis[b] = 1;
update(++ js, a, b, 1);
gs++;
}
} else if (tmp == 2) {
if(gs) {
find_max(1);
gs--;
}
} else if (tmp == 3) {
if(gs) {
find_min(1);
gs--;
}
} else break;
}
cout << e[1].meili << " " << e[1].jiazhi << "\n";
return 0;
}