Fish pond fishing priority queue multi-way merge sort greedy
Introduction to Priority Queue
Greedy thought: Will not look back.
Problem solving thought: enumerate 0-N-1 destination fish pond k,
as shown in the figure, select the number of t (fishing time) from top to bottom, and maximize the sum
#include<iostream>
#include<queue>
using namespace std;
priority_queue<pair<int,int> > q;
int maxx;
int N,T;
int sum[105]; //sum[k] 表示到k鱼塘路程消耗的时间
struct YuT{
//读入鱼塘信息
int beg;
int dt;
}yy[105];
int main()
{
cin>>N;
int a;
for(int i=0;i<N;i++)
{
scanf("%d",&yy[i].beg);
}
for(int i=0;i<N;i++)
{
scanf("%d",&yy[i].dt);
}
for(int i=1;i<N;i++)
{
scanf("%d",&a);
sum[i]=sum[i-1]+a;
}
cin>>T;
int t; //钓鱼的时间
int ans;
for(int k=0;k<N;k++) //枚举目的地
{
// cout<<"k:"<<k<<endl;
ans=0;
while(!q.empty())
{
q.pop();
}
t=T-sum[k];
if(t<=0)
continue;
// cout<<"t:"<<t<<endl;
for(int i=0;i<=k;i++) //把一路的头入队列
{
q.push({
yy[i].beg,i}); //需要记录鱼塘代号，以便入下个队列
}
int near;
while(!q.empty()) //某一路可掉的鱼数归0了
{
pair<int,int> s=q.top(); //每次钓1min鱼
q.pop();
t--;
ans+=s.first; //掉一分钟 钓鱼总数增加
// cout<<s.first<<" "<<s.second<<endl;
if(!t) //时间用完了，跳出循环
{
break;
}
near=s.first-yy[s.second].dt; //计算 下一个入队列的
if(near>0)
q.push({
near,s.second});
}
maxx=max(maxx,ans);
}
cout<<maxx;
return 0;
}