BZOJ2802Warehouse Store题解

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我太菜了，连贪心题都不会写。。。

贪心思路很简单，我们能满足顾客就满足他，如果满足不了，就看之前的顾客中

有没有需求比该顾客多的顾客，如果有的话改为卖给这位顾客会使解更优

所以我们用一个优先队列维护一下所有卖的顾客，然后每此判断一下就好了

# include<iostream>
# include<cstdio>
# include<algorithm>
# include<cstring>
# include<cmath>
# include<queue>
using namespace std;
typedef long long LL;
priority_queue<pair<int,int> > q;
const int mn = 250005;
int n;
int a[mn],b[mn],st[mn],cnt;
bool vis[mn];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
        scanf("%d",&b[i]);
    LL now=0;
    for(int i=1;i<=n;i++)
    {
        now+=a[i];
        if(now-b[i]>=0)
        {
            q.push(make_pair(b[i],i));
            vis[i]=1;
            now-=b[i];
        }
        else if(!q.empty()){
            pair<int,int> tmp=q.top();
            if(tmp.first>b[i] && now+tmp.first-b[i]>=0)
            {
                q.pop();
                vis[tmp.second]=0;
                now=now+tmp.first-b[i];
                vis[i]=1;
                q.push(make_pair(b[i],i));
            }
        }
    }
    for(int i=1;i<=n;i++)
        if(vis[i]) st[++cnt]=i;
    printf("%d\n",cnt);
    for(int i=1;i<=cnt;i++)
        printf("%d ",st[i]);
    return 0;
}