bzoj 4781: [Usaco2017 Open]Paired Up

→题目链接←

每次把最大的牛与最小的牛合并，最终结果一定是最大值最小的

所以我们用优先队列维护一个大根堆，再维护一个小根堆，每次合并更新下答案就好

靠，排个序就好，我是智障，不改了

代码：

#include<iostream>#include<cstring>#include<string>#include<algorithm>#include<queue>#include<cstdio>using namespace std;struct node1{int sum,val,num;friend bool operator < (node1 a,node1 b){return a.val<b.val;}};struct node2{int sum,val,num;friend bool operator < (node2 a,node2 b){return a.val>b.val;}};int n;priority_queue<node1>q1;priority_queue<node2>q2;int ans=0;int main(){scanf("%d",&n);node1 t1;node2 t2;for(int i=1; i<=n; i++){scanf("%d%d",&t1.sum,&t1.val);t2.sum=t1.sum;t2.val=t1.val;t1.num=t2.num=i;q1.push(t1);q2.push(t2);}t1.sum=t2.sum=0;t1.num=0;t2.num=1;while(t1.num!=t2.num){if(t1.sum==0)t1=q1.top(),q1.pop();if(t2.sum==0)t2=q2.top(),q2.pop();if(t1.num==t2.num){ans=max(ans,t1.val*2);break;}int x=min(t1.sum,t2.sum);ans=max(ans,t1.val+t2.val);t1.sum-=x;t2.sum-=x;}printf("%d\n",ans);return 0;}