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1001 Buy and Resell
题目:有1-n个货物,可以在某个点buy,然后在后面的点resell,可以同时买多个,问最大的利润和最小的交易次数。
题解:模拟运算,前 i 天都是可以买的,加入待卖序列(x, 0),对于第 i 天如果最小的待卖的价格比a[i]小,那么说明可以卖,然后将(a[i], 1)加入队列,表示已经卖出,这样模拟一下;
代码:
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 1e5+7;
int n;
ll a[maxn];
struct node{
ll num,index;
node(ll a,ll b)
{
num = a;
index = b;
}
friend bool operator < (node a,node b)
{
if(a.num == b.num) return a.index < b.index;
return a.num > b.num;
}
};
priority_queue <node> q;
int main()
{
//freopen("in.txt", "r", stdin);
int t; scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
ll x = 0, y = 0;
for(int i = 1; i <= n; i++) {
scanf("%lld", &a[i]);
q.push(node(a[i], 0));
node temp = q.top();
if(a[i]>temp.num) {
x += (a[i] - temp.num);
q.pop();
q.push(node(a[i], 1));
}
}
while(!q.empty()) {
if(q.top().index == 1) y++;
q.pop();
}
printf("%lld %lld\n", x, 2ll*y);
}
return 0;
}