JZOJ 5372. 【NOIP2017提高A组模拟9.17】猫

题目大意

对于 (m = [1,lfloor frac n 2 floor])
要求在一个序列中恰好选出 (m) 个不相邻的数使得权值和最大
其中 (1) 的左边是 (n)，(n) 的右边是 (1)

分析

比较经典的贪心
做法
链表记录一个点的前驱后继
然后每次选权值最大的点加入答案，把与它相邻的点标记
然后把它和与它相邻的点缩成一个点，权值为相邻点的权值和减去当前点的权值，加入大根堆
题解

(Code)

#include<cstdio>
#include<queue>
#define LL long long
using namespace std;

const int N = 4e5 + 5;
int n , nxt[N] , pre[N] , vis[N];
LL a[N];

struct node{
	int id; LL v;
	bool operator < (node c) const {return v < c.v;}
};
priority_queue<node> Q;

int main()
{
	freopen("cat.in" , "r" , stdin);
	freopen("cat.out" , "w" , stdout);
	scanf("%d" , &n);
	for(register int i = 1; i <= n; i++)
	{
		scanf("%lld" , &a[i]);
		nxt[i] = (i == n ? 1 : i + 1) , pre[i] = (i == 1 ? n : i - 1);
		Q.push(node{i , a[i]});
	}
	node now; LL ans = 0; int m = n / 2;
	while (m--)
	{
		now = Q.top() , Q.pop();
		while (!Q.empty() && vis[now.id]) now = Q.top() , Q.pop();
		ans += now.v , printf("%lld
" , ans);
		vis[now.id] = vis[nxt[now.id]] = vis[pre[now.id]] = 1;
		a[++n] = a[nxt[now.id]] + a[pre[now.id]] - a[now.id];
		nxt[n] = nxt[nxt[now.id]] , pre[n] = pre[pre[now.id]];
		nxt[pre[n]] = pre[nxt[n]] = n;
		Q.push(node{n , a[n]});
	}
}