HDOJ 1896 Stones 解题报告

题目分类：优先队列+STL

作者：ACShiryu

做题时间：2011-7-18

Stones

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 217    Accepted Submission(s): 107

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

 

题目大意是路上有很多石头，当你遇到奇数序列的石头就把他向前仍，偶数的不动他，如果两个石头一起，先考虑可以仍的比较近的石头仍也就是比较大的石头，这样一直下去，直到前面所有的石头都不可以仍了为止，求最远的石头距离起点多少题目这题用优先队列非常方便.

分析；可以定义一个结构体，分别存储石头现在的位置和能能出去的距离到优先队列中，然后每次取“最小的”，如果取得是偶数个就不动，取得是奇数个就要更新该石头的位置并重新存到优先队列中，直到队列空，输出最后一个石头的位置

数据分析：程序的时间复杂度是O（nlogn）,数据量最大为100,000，不会超时。要特别注意多个石头的x一样的情况,要优先考虑y值最小的那块石头

参考代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
struct Stone{
    int x;    //石头的初始地
    int y;    //石头能扔的最远距离
};
bool operator<( Stone a, Stone b )
{ //重载小于，按照结构体中x小的在队顶，如果x一样，则按照y的最小的在//队顶
    if( a.x== b.x ) return a.y > b.y;  
    return a.x > b.x;   
}  
int main()
{
    int t;
    scanf("%d",&t);//测试数据个数
    while(t--)
    {
        int n;
        int i ;
        priority_queue<Stone>que;     //定义一个Stone成员的优先//队列
        scanf("%d",&n);
        Stone tmp;
        for(i =0;i< n ; i++ )
        {
            scanf("%d%d",&tmp.x,&tmp.y);
            que.push(tmp);//入队
        }
        int sum =1;//判断碰到的是第几个石头的标记
        while(!que.empty())//当队列为空就跳出循环，也就是说再//向前就没有石头可以遇到
        {
        tmp = que.top();//去队顶元素，也就是在后面的所有//石头中第一个碰到的石头
            que.pop();//出对
            if(sum%2)
            {//如果是奇数号石头，则处理，否则不做处理
                tmp . x+=tmp.y;//则向前扔y单位长度
                que.push(tmp);//扔出去的石头入队
            }
            sum++;//石头计数+1
        }
        printf("%d\n",tmp.x);//打印最后一块石头的坐标就是所求//的最远距离
    }
    return 0;
}