Stones
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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.
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.
221 52 421 56 6
1112
题目大意 : 一个人扔石头, 如果第奇数次扔石头, 那么石头被扔出,如果第偶数次扔石头,石头不被扔出 ,可能在一个位置有多快石头,先扔Di小的
代码如下
#include<stdio.h>
#include<queue>
using namespace std;
struct node{
int pi;
int di;
};
bool operator<(node a,node b){ 对结构体使用优先队列
if(a.pi==b.pi) return a.di>b.di;
return a.pi>b.pi;
}
int main(){
int T;
scanf("%d",&T);
while(T--){
int n;
scanf("%d",&n);
priority_queue<node>p;
node s;
int i;
for(i=0;i<n;i++){
scanf("%d%d",&s.pi,&s.di);
p.push(s);
}
int cnt=0;
int Max=0;
while(!p.empty()){
node temp=p.top();
Max=temp.pi;
cnt++;
p.pop();
if(cnt%2==1){
temp.pi+=temp.di;
p.push(temp);
}
}
printf("%d\n",Max);
}
return 0;
}
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