HDU 1896 Stones
Stones
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 374 Accepted Submission(s): 217
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.
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.
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
Recommend
lcy
//这题算做的比较有成就感，呵呵
//将位置p作为第一优先级数据考虑，将D作为第二优先级考虑，都是小堆，这样就可以了，
//187MS，算是很快的了，呵呵
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <vector>
using namespace std;
struct node
{
int p;
int thr;
};
struct cmp
{
bool operator()(const node&a,const node&b)
{
if(a.p==b.p)
{
return a.thr>b.thr;
}
return a.p>b.p;
}
};
int main()
{
int T;
scanf("%d",&T);
int n,k,max;
node t;
while(T--)
{
scanf("%d",&n);k=1;max=0;
priority_queue<node,vector<node>,cmp> Q;
while(n--)
{
scanf("%d%d",&t.p,&t.thr);
Q.push(t);
}
while(!Q.empty())
{
if(k%2)
{ k++;
t=Q.top();
Q.pop();
t.p=t.p+t.thr;
if(t.p>max) max=t.p;
Q.push(t);
}
else
{ k++;
Q.pop();
}
}
printf("%d\n",max);
}
return 0;
}
#include <stdio.h>
#include <string.h>
#include <queue>
#include <vector>
using namespace std;
struct node
{
int p;
int thr;
};
struct cmp
{
bool operator()(const node&a,const node&b)
{
if(a.p==b.p)
{
return a.thr>b.thr;
}
return a.p>b.p;
}
};
int main()
{
int T;
scanf("%d",&T);
int n,k,max;
node t;
while(T--)
{
scanf("%d",&n);k=1;max=0;
priority_queue<node,vector<node>,cmp> Q;
while(n--)
{
scanf("%d%d",&t.p,&t.thr);
Q.push(t);
}
while(!Q.empty())
{
if(k%2)
{ k++;
t=Q.top();
Q.pop();
t.p=t.p+t.thr;
if(t.p>max) max=t.p;
Q.push(t);
}
else
{ k++;
Q.pop();
}
}
printf("%d\n",max);
}
return 0;
}
