ZOJ 3230 Solving the Problems（数学 优先队列啊）

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3230

Programming is fun, Aaron is addicted to it. In order to improve his programming skill, he decides to solve one programming problem per day. As you know, different problems have different properties, some problems are so difficult that there are few people can solve it, while some problems are so easy that almost everyone is able to tackle it.

Programming skill can be measured by an integer p. And all problems are described by two integers a_i and b_i. a_i indicates that if and only if P >= a_i, you can solve this problem. b_i indicates that after you solve this problem, your programming skill can be increased by b_i.

Given the initial programming skill p of Aaron, and the information of each problem, Aaron want to know the maximal programming skill he can reach after m days, can you help him?

Input

Input consists of multiple test cases (less than 40 cases)!

For each test case, the first line contains three numbers: n, m, p (1 <= n <= 100000, 1 <= m <= n, 1 <= p <= 10000), n is the number of problems available for Aaron,m, p as mentioned above.

The following n lines each contain two numbers: a_i and b_i (1 <= a_i <= 10000, 1 <= b_i <= 10000) describe the information of the i-th problem as memtioned above.

There's a blank line between consecutive cases.

Output

For each case, output the maximal programming skill Aaron can reach after m days in a line.

Sample Input

2 2 1
1 2
7 3

3 1 2
1 2
2 3
3 4

Sample Output

3
5

---

Author: ZHOU, Yilun
Source: ZOJ Monthly, July 2009

题意：

给出Aaron 的初始的做题能力值，然后给出N道题：Aaron 解决这道题至少须要多少能力值a。Aaron 攻克了这道题他能添加多少能力值b！

求M天后Aaron 的最大能力值是多少！

PS:

运用优先队列。分别对a和b排序一下！

代码例如以下：

//#pragma warning (disable:4786)
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <climits>
#include <ctype.h>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <deque>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
const double eps = 1e-9;
//const double pi = atan(1.0)*4;
const double pi = 3.1415926535897932384626;
const double e = exp(1.0);
#define INF 0x3f3f3f3f
//#define INF 1e18
//typedef long long LL;
//typedef __int64 LL;
#define ONLINE_JUDGE
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
struct node1
{
    int a, b;
    node1() {}
    node1(int _a, int _b): a(_a), b(_b) {}
    bool operator < (const node1 & x) const
    {
        return a > x.a;//最小值优先
    }
};
struct node2
{
    int a, b;
    node2() {}
    node2(int _a, int _b): a(_a), b(_b) {}
    bool operator < (const node2 & x) const
    {
        return b < x.b;//最大值优先
    }
};
int n, m, p;

int main()
{
    int a, b;
    while(~scanf("%d%d%d", &n, &m, &p))
    {
        priority_queue<node1> q1;
        priority_queue<node2> q2;

        for(int i = 0; i < n; i++)
        {
            scanf("%d%d", &a, &b);
            q1.push(node1(a, b));
        }
        int maxx_p = p;
        for(int i = 0; i < m; i++)
        {
            while (!q1.empty())
            {
                int tt_a = q1.top().a;
                int tt_b = q1.top().b;
                if(tt_a <= maxx_p)
                {
                    q2.push(node2(tt_a, tt_b));
                    q1.pop();
                }
                else
                {
                    break;
                }
            }
            if (q2.empty())
            {
                break;
            }
            maxx_p += q2.top().b;
            q2.pop();
        }
        printf("%d
", maxx_p);
    }
    return 0;
}
/*
3 1 2
1 2
2 3
3 4
3 2 2
1 2
2 3
3 4
*/