没办法了只能特判了
by courtesan @ 2019-02-14 14:48:02
```
//思路：到达每个站点的最小耗油量都等于在它一箱油的距离内
//所有站点的最小耗油量+到达该站点的耗油量 中最小的一个
#include<bits/stdc++.h>
using namespace std;
typedef long double ld;
struct station
{
ld cost;
ld pri;//分别代表从起点到该站的距离和该站的油价
};
vector<station> all(8);
vector<ld> cost(8,1000000);
ld d1,c,d2,p,n;
int main()
{
cin>>d1>>c>>d2>>p>>n;
all[0].cost=0,all[0].pri=p;
all[n+1].cost=d1,all[n+1].pri=10000;//把起点和终点看作两个加油站并设置好
for(int i=1;i<=n;i++)
cin>>all[i].cost>>all[i].pri;
cost[0]=0;
for(int i=1;i<=n+1;i++)
{if(all[i].cost-all[i-1].cost>c*d2)
{cout<<"No Solution";exit(0);
}
for(int j=0;j<i;j++)
if(all[i].cost-all[j].cost<=c*d2)
cost[i]=min(cost[i],cost[j]+(all[i].cost-all[j].cost)/d2*all[j].pri);//转移方程
}
cout<<fixed<<setprecision(2)<<cost[n+1];
}
```
by courtesan @ 2019-02-14 14:54:58
@[courtesan](/space/show?uid=179441) 我我我
by ycyaw @ 2019-02-28 18:34:20
捞捞沉帖
by ycyaw @ 2019-02-28 18:34:42
计算没有很大的问题，你应该是贪心出来问题，之前我也出现过这种问题
by whwh @ 2019-06-07 07:57:33
+1
by 耶梦加得 @ 2019-07-09 16:21:39
+1
by z_y_z @ 2019-07-10 19:38:21
+1
by ysmlwudia @ 2019-08-12 10:12:17