题目地址：http://codeforces.com/contest/441/problem/B
B. Valera and Fruits
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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 than v 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?
Input
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.
Output
Print a single integer — the maximum number of fruit that Valera can collect.
Examples
input
2 3
1 5
2 3
output
8
input
5 10
3 20
2 20
1 20
4 20
5 20
output
60
Note
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.
题目大意：
果农摘果子，一共有n棵树，农夫每天可以摘v个果子
每个果子保存时间为两天，两天后未摘的就不能吃了
接下来n行表示某天的果子数量 //第几天与n没有关系！！！
思路：
优先考虑前一天
注意第几天与n没有关系！！！
用一个max保存最大天数然后遍历即可
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int main()
{
int n,v;
scanf("%d%d",&n,&v);
int a[100000]={0};
int max=0;
for(int i=1;i<=n;i++)
{
int item,item2;
scanf("%d",&item);
scanf("%d",&item2);
if(item2<=0)
item2=0;
a[item]+=item2;
if(item>max)
max=item;
}
if(v<=0)
{
printf("0\n");
return 0;
}
int sum=0;
int flag=0;
for(int i=1;i<=max;i++)
{
if(a[i-1]>0)
{
if(a[i-1]>=v)
{
sum+=v;
continue;
}
else
{
sum+=a[i-1];
if(v-a[i-1]>=a[i])
{
sum+=a[i];
a[i]=0;
continue;
}
else
{
sum+=v-a[i-1];
a[i]-=(v-a[i-1]);
}
}
}
else
{
if(a[i]<=v)
{
sum+=a[i];
a[i]=0;
}
else
{
sum+=v;
a[i]-=v;
}
}
}
if(a[max]>0)
{
if(a[max]>=v)
sum+=v;
else
sum+=a[max];
}
printf("%d\n",sum);
return 0;
}