HDU 1896

最新推荐文章于 2020-11-04 06:55:19 发布

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本文介绍了一个基于石头的游戏模拟问题，主人公Sempr会在每天行走过程中遇到不同的石头，并根据石头的奇偶性进行处理。游戏的目标是计算Sempr走过后最远石头与起点的距离。

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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

#include <cstdio>
#include <queue>
using namespace std;
#define M 110000

struct node
{
    int pi, di;
    friend bool operator <(node t1, node t2)//队列以距离排序
    {
        if(t1.pi == t2.pi)//距离相等时先看到扔的远的
        {
            return t1.di > t2.di;
        }
        return t1.pi > t2.pi;
    }
};
node no[M], no2;
int main()
{
    int t, n;
    scanf("%d", &t);
    while(t--)
    {
        priority_queue<node> q;
        scanf("%d", &n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d", &no[i].pi, &no[i].di);
            q.push(no[i]);
        }
        int jo = 1;
        int a2, a1, ans;
        while(!q.empty())
        {
            no2 = q.top();
            ans = no2.pi;
            q.pop();
            if(jo % 2)//判断奇偶
            {
                no2.pi = no2.pi + no2.di;
                q.push(no2); 
            }
            jo++;
        }
        printf("%d\n", ans);
    }

    return 0;
}