1 second
256 megabytes
standard input
standard output
Valera loves his garden, where n fruit trees grow.
This year he will enjoy a great harvest! On the i-th tree bi fruit
grow, they will ripen on a day number ai.
Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and
day ai + 1 (all
fruits that are not collected in these two days, become unfit to eat).
Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more thanv fruits.
The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?
The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) —
the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.
Next n lines contain the description of trees in the garden. The i-th
line contains two space-separated integers ai and bi(1 ≤ ai, bi ≤ 3000) —
the day the fruits ripen on the i-th tree and the number of fruits on the i-th
tree.
Print a single integer — the maximum number of fruit that Valera can collect.
2 3
1 5
2 3
8
5 10
3 20
2 20
1 20
4 20
5 20
60
In the first sample, in order to obtain the optimal answer, you should act as follows.
- On the first day collect 3 fruits from the 1-st
tree. - On the second day collect 1 fruit from the 2-nd
tree and 2 fruits from the 1-st
tree. - On the third day collect the remaining fruits from the 2-nd tree.
In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.
题目链接：http://codeforces.com/contest/441/problem/B
题目大意：有n棵果树和每天最大的水果採摘量v。每棵果树有相应的採摘日期day，仅仅能在day和day+1两天採摘，求终于水果的最大採摘量。
解题思路：对每棵果树的採摘日期从小到大排序，日期靠前的先採摘。
代码例如以下：
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
int const maxn=300005;
struct A
{
int d,b;
}a[3005];
int cmp(A p,A q)
{
return p.d<q.d;
}
int main()
{ int n,v,ans=0;
scanf("%d%d",&n,&v);
for(int i=1;i<=n;i++)
{
scanf("%d%d",&a[i].d,&a[i].b);
}
sort(a+1,a+n+1,cmp);
for(int day=1;day<=maxn;day++)
{
int cur=0,p=0; //cur为每天的採摘量
for(int i=1;i<=n;i++)
{
if(a[i].d>day) //假设还未到达当前果树的日期，后面果树的日期也未到达，结束当前循环
break;
if(a[i].d!=day&&a[i].d+1!=day)
continue;
if(a[i].b>0) //在规定的两天日期能够採摘
{
p=1;
if(a[i].b+cur<=v) //a[i].b表示当前果树上果实个数。假设採摘数量会有更新
{
cur+=a[i].b;
a[i].b=0;
}
else
{
a[i].b-=(v-cur);
cur=v;
break;
}
}
}
ans+=cur;
if(a[n].d+1<day)
break;
}
printf("%d\n",ans);
return 0;
}