杭电1896（题意详解+代码）（优先队列）之Stones

Problem Description

Because of the wrong status of the bicycle, Sempr begin to walk east to west every morning and walk back every evening. Walking may cause a little tired, so Sempr always play some games this time.
There are many stones on the road, when he meet a stone, he will throw it ahead as far as possible if it is the odd stone he meet, or leave it where it was if it is the even stone. Now give you some informations about the stones on the road, you are to tell me the distance from the start point to the farthest stone after Sempr walk by. Please pay attention that if two or more stones stay at the same position, you will meet the larger one(the one with the smallest Di, as described in the Input) first.

 

Input

In the first line, there is an Integer T(1<=T<=10), which means the test cases in the input file. Then followed by T test cases.
For each test case, I will give you an Integer N(0<N<=100,000) in the first line, which means the number of stones on the road. Then followed by N lines and there are two integers Pi(0<=Pi<=100,000) and Di(0<=Di<=1,000) in the line, which means the position of the i-th stone and how far Sempr can throw it.

 

Output

Just output one line for one test case, as described in the Description.

 

Sample Input

2
2
1 5
2 4
2
1 5
6 6

 

Sample Output

11
12

题目大意：
一个人在路上从0位置出发，从左往右走，第奇数次遇到石头就把它往右扔出去，偶数次遇到石头就不管它，当他在某个位置遇到两个或两个以上的石头，最先遇到的将是d值最小的那个，然后依次按d值从小到大的书序遇到该位置上的石头，计算 他扔的 最后那块石头离起点的距离？？？   看一下下面的案例就懂了

分析：：
4
1 5
2 4
3 3
4 2

1 2 3 4  >> 2 3 4 6  >> (2) 3 4 6 >> 4   6<2>  >> (4)  6<2>  >> 6  9 >> (6) 9 >> 12
5 4 3 2     4 3 2 5      4  3 2 5    4   3,5       4   3,5      5  3     5  3   

     
5
1 5
2 4
3 3
7 8
8 100
输出 208

建立坐标轴:                                                                    
初始状态： 1  2  3  7  8  >>> 一、 2  3  6  7  8  >>> 二、 (2)  3  6  7  8  >>> 三、6<2>   7  8  >>> 四、（6） 6  7    8   >>> ...
	 5  4  3  8  100         4  3  5  8  100             3  5  8  100        5、3   7  100         3   5  8   100     
     .... >>> 五、7  8     11  >>> 六、 (7)  8   11  >>> 七、  11   108   >>>八、  (11)  108  >>> 九、 208
                 8  100    5           8   100  5            5    100             5   100

解释说明：
1，   ...>>>...>>>...表示位置信息Pi，下面一排数表示对应位置的石头能扔的距离Di 
2，   一、 表示第一次遇到石头操作后的结果，  二、 表示第二次遇到石头操作后的结果
3，   (n) 表示n这个石头被跳过了，所以有（）的那次操作，前面的中文数字必定是偶数  
4，   n<m> 表示n位置有m个石头        
5，    n,m  表示当前为有两块石头，并且两块石头可以扔的距离分别为n和m
T组测试案例，n表示石头数量， 每组有n对数，Pi,Di； 分别表示第i块石头所在位置，和第i块石头能扔的距离


思路：涉及 动态排序+最优问题 ，故用 优先队列 解决

AC代码如下： 
#include <iostream>
#include <cstdio>
#include <queue>
using namespace std;

struct Node  //石头信息
{
    int p;
    int d;
};

struct cmp
{
    bool operator() (const Node a,const Node b)const   //a优先级小于b，返回true
    {
        if(a.p==b.p) return a.d>b.d;
        else return a.p>b.p;
    }
};

int main()
{
    int t,n;
    cin>>t;
    while(t--)
    {
        priority_queue <Node,vector<Node>,cmp> pq;  //定义优先队列
        Node node;
        cin>>n;
        while(!pq.empty()) pq.pop();    //清空队列
        for(int i=0;i<n;i++)
        {
            cin>>node.p>>node.d;     //入队列
            pq.push(node);
        }
        int flag=1;                 //记录遇到石头的次数
        while(!pq.empty())
        {
            node=pq.top(); pq.pop();
            if(flag%2==1)
            {
                node.p+=node.d;
                pq.push(node);
            }
            flag++;
        }
        cout<<node.p<<endl;
    }
    return 0;
}