HDU - 1896 Stones（优先队列）

最新推荐文章于 2022-01-23 16:24:15 发布

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最新推荐文章于 2022-01-23 16:24:15 发布

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本文介绍了一个关于石头位置变化的算法题，并提供了详细的解题思路和代码实现。题目中，Sempr 每天沿着一条路来回行走，遇到奇数编号的石头会将其向前扔出一定距离，而偶数编号的石头则保持不动。任务是求出最远石头的位置。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1896点击打开链接

Stones

Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2885 Accepted Submission(s): 1845

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

Sample Output

  
   11
12

优先队列距离近的先出来判断

注意奇数偶数不同情况

#include <iostream>
#include <queue>
#include <stdio.h>
#include <stdlib.h>
#include <stack>
#include <limits>
#include <string>
#include <string.h>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
using namespace std;
struct xjy
{
    int now;
    int step;
    bool operator < ( const xjy &r)const
    {
        if(now==r.now)
            return step>r.step;
        return now>r.now;
    }
};
priority_queue<xjy>q;
int main()
{
    int t=0;
    scanf("%d",&t);
    while(t--)
    {
        while(!q.empty())
            q.pop();
        int n=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            {
                xjy mid;
                scanf("%d%d",&mid.now,&mid.step);
                q.push(mid);
            }
        int num=1;
        int ans=0;
        while(!q.empty())
        {
            xjy mid;
            mid=q.top();
            q.pop();
            if(num&1)
            {
                ans=max(ans,mid.now);
                mid.now+=mid.step;
                q.push(mid);
                num++;
            }
            else
            {
                ans=max(ans,mid.now);
                num++;
            }
        }
        printf("%d\n",ans);
    }
}