hdu 1896 优先队列
2017-07-12 21:38
351 查看
题目:
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2757 Accepted Submission(s): 1778
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
Recommend
lcy | We have carefully selected several similar problems for you: 1892 1899 1895 1894 1897
代码:
优先队列比较函数的另外两种姿势:
http://www.cnblogs.com/hxsyl/archive/2012/08/30/2662955.html http://blog.csdn.net/liangzhaoyang1/article/details/52876788
Stones
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 2757 Accepted Submission(s): 1778
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
Recommend
lcy | We have carefully selected several similar problems for you: 1892 1899 1895 1894 1897
代码:
#include<stdio.h> #include<iostream> #include<queue> #include<algorithm> using namespace std; struct node{ int p,d; node(int pp,int dd){ p=pp; d=dd; } bool operator< (const node & a)const{ if(p==a.p) return d>a.d;///////>>>>>>>>>>>>>> else return p>a.p;/////............... } }; int t,n,a,b; priority_queue<node> qu; int main(){///265MS 3480K scanf("%d",&t); while(t--){ scanf("%d",&n); while(!qu.empty()) qu.pop(); for(int i=1;i<=n;++i) scanf("%d%d",&a,&b),qu.push(node(a,b)); int num=0,ans=0; //node tmp; while(!qu.empty()){ node tmp=qu.top(); qu.pop(); num++; //cout<<num<<" "<<tmp.p<<" "<<tmp.d<<endl; if(num%2) qu.push(node(tmp.p+tmp.d,tmp.d)); else ans=max(ans,tmp.p); } printf("%d\n",ans); } return 0; }
优先队列比较函数的另外两种姿势:
http://www.cnblogs.com/hxsyl/archive/2012/08/30/2662955.html http://blog.csdn.net/liangzhaoyang1/article/details/52876788
相关文章推荐
- HDU 1896 Stones ——STL 优先队列
- HDU 1896 Stones (优先队列的应用)
- hdu 1896 优先队列的应用
- hdu 杭电1896 Stones【优先队列】
- HDU - 1896 : Stones(优先队列、有序对)
- hdu 1896 优先队列的应用
- HDU 1896 Stones --优先队列+搜索
- 【HDU]-1896-Stones(优先队列,好)
- HDU - 1896 Stones —— 优先队列——贪心
- hdu 1896(2013.9.15周赛D题)优先队列
- HDU 1896 Stones 【优先队列】
- HDU 1896 Stones --优先队列+搜索
- hdu 1896 Stones(优先队列 Dijkstr)
- hdu diy Developing School's Contest 2012-6 by SYU 优先队列排序问题
- hdu 2680 dijkstra 使用优先队列优化
- HDU 4006 The kth great number (堆实现优先队列)
- hdu 2757 优先队列
- 同样优先队列的广搜题 hdu 4198
- HDU_1242——二维空间搜索,使用优先队列BFS
- HDU 1180 诡异的楼梯 -- 优先队列