hdu1896(模拟+优先级队列)

Stones

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 659    Accepted Submission(s): 389

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

221 52 421 56 6

 

Sample Output

1112
 
本题是个模拟题，但若果直接模拟超内存或超时，可以选择用优先级队列，先按位置从小到大排序，若位置相同则按距离从小到大排序，每次去队头元素，若为取出的第奇数个，则出队列后修改位置后新的元素入队列；若取出的为第偶数个，则直接出队列。
#include<iostream>#include<queue>#include<algorithm>using namespace std;struct stone{int pos,dis;}sto[100000+10];struct cmp{bool operator()(stone a,stone b){if(a.pos!=b.pos)return a.pos>b.pos;else return a.dis>b.dis;}};int main(){int cas,i,n,tot,k;cin>>cas;while(cas--){scanf("%d",&n);priority_queue<stone,vector<stone>,cmp>qq;for(i=0;i<n;i++){scanf("%d%d",&sto[i].pos,&sto[i].dis);qq.push(sto[i]);}k=1;stone tmp;while(!qq.empty()){tmp=qq.top();//if(1==k)//tot+=tmp.pos;qq.pop();if(k%2){//cout<<tmp.pos<<"&&&&"<<tmp.dis<<endl;    //tot+=tmp.dis;//cout<<tot<<"****"<<endl;tmp.pos+=tmp.dis;qq.push(tmp);k++;}elsek++;}//tot+=tmp.dis;printf("%d\n",tmp.pos);}return 0;}