HDU-1896 Stones

http://acm.hdu.edu.cn/showproblem.php?pid=1896

题意：一个人从0开始走起，遇到偶数个石头就踢。要是同一位置有多个石头，则先扔最重的石头（也就是扔的最近的那个石头），要你求扔的石头离初始位置的最大距离。

Stones

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1231    Accepted Submission(s):
759

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

 

 

Author

Sempr|CrazyBird|hust07p43

 

Source

HDU
2008-4 Programming Contest

 

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 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 using namespace std;
 struct node
 {
      friend bool operator<(node n1,node n2)
     {

           if(n1.p>n2.p)//小的优先级大。
                  return ;
            else
             {
                 if(n1.p==n2.p&&n1.d>n2.d)
                 {
                     return ;
                 }
                 else
                    return ;
             }

     }
     int p;
     int d;
 };
 int main()
 {
     int t,f;
     scanf("%d",&t);
     while(t--)
     {
         int n,i;
        scanf("%d",&n);
         struct  node a;
         priority_queue<node>q;
        for(i=;i<n;i++)
          {
              scanf("%d%d",&a.p,&a.d);
              q.push(a);
          }
          int s=;
          int ans=;
          while(!q.empty())
          {
              s++;
              a=q.top();
            //  printf("p1=%d,d1=%d\n",a.p,a.d);
              q.pop();
              if(s%==)
              {
               // printf("p=%d,d=%d\n",a.p,a.d);
               // ans+=a.p+a.d;
                a.p=a.p+a.d;
              //  f=a.d;
               q.push(a);
              }

           }

         printf("%d\n",a.p);
     }
     return ;
 }