标签:错误 ogr ace lse ons long operator main http
原题链接
考察:贪心
错误思路:
??对于每个任务,按d,a顺序排序,如果不能按时完成就付钱使得按时完成.
思路:
??不一定要压当前任务的时间,我们可以压花费更小的任务时间,使得超时任务按时完成.
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <queue>
using namespace std;
const int N = 100010;
typedef long long LL;
struct Node{
int a,b,d;
}node[N];
struct cmp2{
bool operator()(Node n,Node m){
if(m.d==n.d) return m.a>n.a;
return n.d<m.d;
}
};
struct cmp{
bool operator()(Node n,Node m){
return n.a<m.a;
}
};
int n;
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
int a,b,d;
scanf("%d%d%d",&a,&b,&d);
node[i].a = a,node[i].b = b,node[i].d = d;
}
double ans = 0;
sort(node+1,node+n+1,cmp2());
priority_queue<Node,vector<Node>,cmp> q;
LL now = 0;
for(int i=1;i<=n;i++)
{
now+=(LL)node[i].b;
q.push(node[i]);
while(q.size()&&now>node[i].d)
{
Node it = q.top();
q.pop();
if(now-node[i].d>=it.b)
{
ans+=1.0*it.b/it.a;
now-=it.b;
}else{
ans+=(now-node[i].d)*1.0/it.a;
it.b-=now-node[i].d;
now-=now-node[i].d;
q.push(it);
}
}
}
printf("%.2lf\n",ans);
return 0;
}
The lazy programmer POJ - 2970
标签:错误 ogr ace lse ons long operator main http
原文地址:https://www.cnblogs.com/newblg/p/14901422.html