Knight Tournament (set)
Hooray! Berl II, the king of Berland is making a knight tournament. The king has already sent the message to all knights in the kingdom and they in turn agreed to participate in this grand event.
As for you, you're just a simple peasant. There's no surprise that you slept in this morning and were late for the tournament (it was a weekend, after all). Now you are really curious about the results of the tournament. This time the tournament in Berland went as follows:
- There are n knights participating in the tournament. Each knight was assigned his unique number — an integer from 1 to n.
- The tournament consisted of m fights, in the i-th fight the knights that were still in the game with numbers at least li and at most ri have fought for the right to continue taking part in the tournament.
- After the i-th fight among all participants of the fight only one knight won — the knight number xi, he continued participating in the tournament. Other knights left the tournament.
- The winner of the last (the m-th) fight (the knight number xm) became the winner of the tournament.
You fished out all the information about the fights from your friends. Now for each knight you want to know the name of the knight he was conquered by. We think that the knight number b was conquered by the knight number a, if there was a fight with both of these knights present and the winner was the knight number a.
Write the code that calculates for each knight, the name of the knight that beat him.
Input
The first line contains two integers n, m (2 ≤ n ≤ 3·105; 1 ≤ m ≤ 3·105) — the number of knights and the number of fights. Each of the following m lines contains three integers li, ri, xi (1 ≤ li < ri ≤ n; li ≤ xi ≤ ri) — the description of the i-th fight.
It is guaranteed that the input is correct and matches the problem statement. It is guaranteed that at least two knights took part in each battle.
Output
Print n integers. If the i-th knight lost, then the i-th number should equal the number of the knight that beat the knight number i. If the i-th knight is the winner, then the i-th number must equal 0.
题解:用set来模拟
代码:
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
using namespace std;
int a[1000005];
int main()
{
set<int>s;
set<int>::iterator it;
int n,m;
cin>>n>>m;
s.clear();
for(int t=1;t<=n;t++)
{
s.insert(t);
}
int l,r,x;
for (int t=1;t<=m;t++)
{
x=s.size();
scanf("%d%d%d",&l,&r,&x);
it=s.lower_bound(l);
int h;
while (it!=s.end() && *it<=r)
{
h=*it;
it++;
if (h!=x)
{
a[h]=x;
s.erase(h);
}
}
}
printf("%d",a[1]);
for(int t=2;t<=n;t++)
{
printf(" %d",a[t]);
}
return 0;
}
Knight Tournament (set)的更多相关文章
- D - Knight Tournament(set)
Problem description Hooray! Berl II, the king of Berland is making a knight tournament. The king has ...
- CodeForce 356A Knight Tournament(set应用)
Knight Tournament time limit per test 3 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #207 (Div. 1) A. Knight Tournament (线段树离线)
题目:http://codeforces.com/problemset/problem/356/A 题意:首先给你n,m,代表有n个人还有m次描述,下面m行,每行l,r,x,代表l到r这个区间都被x所 ...
- CodeForces - 357C Knight Tournament 伪并查集(区间合并)
Knight Tournament Hooray! Berl II, the king of Berland is making a knight tournament. The king has a ...
- Knight Tournament 合并区间
Hooray! Berl II, the king of Berland is making a knight tournament. The king has already sent the me ...
- codeforces 357C Knight Tournament(set)
Description Hooray! Berl II, the king of Berland is making a knight tournament. The king has already ...
- HDU 1372 Knight Moves (bfs)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1372 Knight Moves Time Limit: 2000/1000 MS (Java/Othe ...
- uva439 - Knight Moves(BFS求最短路)
题意:8*8国际象棋棋盘,求马从起点到终点的最少步数. 编写时犯的错误:1.结构体内没构造.2.bfs函数里返回条件误写成起点.3.主函数里取行标时未注意书中的图. #include<iostr ...
- poj2243 Knight Moves(BFS)
题目链接 http://poj.org/problem?id=2243 题意 输入8*8国际象棋棋盘上的两颗棋子(a~h表示列,1~8表示行),求马从一颗棋子跳到另一颗棋子需要的最短路径. 思路 使用 ...
随机推荐
- Gym 100801B Black and White(构造)
题意:给定X,Y,分别表示由'.'和'@'组成的连通块的个数. 思路:假如X<Y,我们用两部分来构造这个结果,第一部分由一个'.'连通块和Y-(X-1)割'@'连通块组成,第二个部分由X-1个' ...
- 2017-2018-1 20179215《Linux内核原理与分析》第七周作业
一.实验部分:分析Linux内核创建一个新进程的过程. [第一部分] 根据要求完成第一部分,步骤如下: 1. 首先进入虚拟机,打开终端,这命令行依次敲入以下命令: cd LinuxKernel ...
- 霍夫变换Hough
http://blog.csdn.net/sudohello/article/details/51335237 霍夫变换Hough 霍夫变换(Hough)是一个非常重要的检测间断点边界形状的方法.它通 ...
- javascript:function 函数声明和函数表达式 详解
函数声明(缩写为FD)是这样一种函数: 有一个特定的名称 在源码中的位置:要么处于程序级(Program level),要么处于其它函数的主体(FunctionBody)中 在进入上下文阶段创建 影响 ...
- 随机数 while循环 do while循环 for循环
1.随机数 arc4random() 返回一个随机数 如果要随机[a,b]范围内的随机数 arc4random() % (b - a + 1) + a ; 2.break 跳出本次循 ...
- Poj 2299 Ultra-QuickSort(归并排序求逆序数)
一.题意 给定数组,求交换几次相邻元素能是数组有序. 二.题解 刚开始以为是水题,心想这不就是简单的冒泡排序么.但是毫无疑问地超时了,因为题目中n<500000,而冒泡排序总的平均时间复杂度为, ...
- lua 函数调用 -- 闭包详解和C调用
转自:http://www.cnblogs.com/ringofthec/archive/2010/11/05/luaClosure.html 这里, 简单的记录一下lua中闭包的知识和C闭包调用 前 ...
- C#设计模式(11)——外观模式
一.概念 外观模式提供了一个统一的接口,用来访问子系统中的一群接口.外观定义了一个高层接口,让子系统更容易使用.使用外观模式时,我们创建了一个统一的类,用来包装子系统中一个或多个复杂的类,客户端可以直 ...
- C#设计模式(10)——组合模式
一.概念 组合模式有时候又叫做部分-整体模式,它使我们树型结构的问题中,模糊了简单元素和复杂元素的概念,客户程序可以向处理简单元素一样来处理复杂元素,从而使得客户程序与复杂元素的内部结构解耦. 二.组 ...
- WCF服务用户名密码访问
有2种方式, 第一直接在程序中指定用户名密码,配置调用 private void BtnSearch_Click(object sender, EventArgs e) { try { var cli ...