CSU 1554 SG Value —— 思维

题目链接：http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1554

Description

The SG value of a set (multiset) is the minimum positive integer that could not be constituted of the number in this set.
You will be start with an empty set, now there are two opertions:
1. insert a number x into the set;
2. query the SG value of current set.

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1e5) -- the total number of opertions.
The next N lines contain one opertion each.
1 x means insert a namber x into the set;
2 means query the SG value of current set.

Output

For each query output the SG value of current set.

Sample Input

5
2
1 1
2
1 1
2

Sample Output

1
2
3

题解：

设当前的 SG value 为now，插入的值为x。

1.如果x>now， 那么now的值不会改变。把这个数放进集合中；

2.如果x<=now， 那么now += x， 由于此次now的增大，在集合中小于等于now的数，还能够使now继续增大。使得now增大之后，这个数应该从集合中删去。

（可以用优先队列或multiset维护这个集合）

学习之处：

升序的优先队列： priority_queue<LL, vector<LL>, greater<LL> >q;

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 20;

priority_queue<LL, vector<LL>, greater<LL> >q;//优先队列的升序写法
LL n, op, x, now;

void init()
{
    while(!q.empty()) q.pop();
    now = 1;
}

void solve()
{
    while(n--)
    {
        scanf("%lld",&op);
        if(op==1)
        {
            scanf("%lld",&x);
            if(x>now)
            {
                q.push(x);
                continue;
            }

            now += x;
            while(!q.empty() && now>=q.top())
            {
                now += q.top();
                q.pop();
            }
        }
        else printf("%lld\n", now);
    }
}

int main()
{
    while(scanf("%lld",&n)!=EOF)
    {
        init();
        solve();
    }
    return 0;
}