codeforces-441B-Valera and Fruits【暴力】

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codeforces-441B-Valera and Fruits

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                    time limit per test1 second     memory limit per test256 megabytes

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.

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.

题目链接：cf-441B

题目大意：有一个果园里有很多树，上面有很多果实，为了不然成熟的果实腐烂，必须在两天之内收集起来。给出果园有的树，以及该树上的果实个数，工人每天可以采集的上限，求出这段时间之后，能收集到的最大值

题目思路：直接暴力就可以了。用数组ans[]存储每天收集的果实数，优先收集前一天的果实，如果还能收集收集今天的果实，剩下的给ans[i+1]；如果不能收集果实，将今天的果实留给下一天ans[i+1]，需要注意的是前一天的果实腐烂不要。

题目很坑的是，可以同一天有多个果实树成熟。

以下是代码：

#include <vector>#include <map>#include <set>#include <algorithm>#include <iostream>#include <cstdio>#include <cmath>#include <cstdlib>#include <string>#include <cstring>using namespace std;int p[3005];int ans[5000];int main(){    int n,v;    cin >> n >> v;    int up = 0;    for (int i = 1; i <= n; i++)    {        int x,y;        cin >> x >> y;        p[x] += y;        up = max(up,x);    }    for(int i = 1; i <= up + 1; i++)    {        int ret = p[i] + ans[i];        int l = ans[i];        if (ret > v)        {                       ans[i] = v;            ans[i + 1] = ret - v;               if (l > v) ans[i + 1] -= l - v;        }                   else  ans[i] = ret;    }    int last = 0;    for (int i = 1; i <= up + 1; i++)    {        last += ans[i];    }       cout << last << endl;    return 0;}