HDU 1896 Stones 优先队列

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探讨一个有趣的石头游戏算法问题，主人公在行走过程中依据石头的奇偶性决定是否投掷，旨在找到最远石头的位置。使用优先队列实现高效算法。

Stones

Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3067 Accepted Submission(s): 1992

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

Author

Sempr|CrazyBird|hust07p43

Source

HDU 2008-4 Programming Contest

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lcy

Statistic | Submit | Discuss | Note 传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1896

题意：熊孩子无聊扔石头，见到的第奇数个石头就扔，第偶数个石头留在原地，如果有多个石头在同一个地方，先看到大的（可扔距离小的）

思路：学校集训时的STL联系，首先想到用STL的优先队列，按距离从小到大排序，距离相等从大到小排序，每走一步就把队首石头拿出来操作；

AC代码：

#include<cstdio>
#include<queue>
using namespace std;

struct stone{
    int pi;
    int di;
};

struct cmp1{
    bool operator()(stone& a,stone& b){
        if(a.pi==b.pi) return a.di>b.di;
        else return a.pi>b.pi;
    }
};

priority_queue<stone ,vector<stone>,cmp1> q;

int m[100050];

void solve(){
    for(int i=1;;i++){
        stone t;
        if(i%2==1){
            t = q.top();
            q.pop();
            t.pi += t.di;
            q.push(t);
        }
        if(i%2==0) q.pop();
        if(q.empty()){
            printf("%d\n",t.pi);
            return ;
        }
    }
}

int main(){
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        int pi,di;
        for(int i=0;i<n;i++){
            scanf("%d%d",&pi,&di);
            stone s;
            s.pi=pi;
            s.di=di;
            q.push(s);
        }
        solve();
    }
    return 0;
}