B. Light bulbs（2019 ICPC上海站）

最新推荐文章于 2022-07-18 10:10:40 发布

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本文探讨了一种涉及大量灯泡的状态翻转算法问题。通过分析初始关闭的灯泡在一系列连续子集翻转操作后的最终亮灯数量，提供了一个简洁的解决方案。代码示例展示了如何使用排序和计数策略来高效解决此问题。

题目描述：

There are NNN light bulbs indexed from 000 to N−1N-1N−1. Initially, all of them are off.A FLIP operation switches the state of a contiguous subset of bulbs. FLIP(L,R)FLIP(L, R)FLIP(L,R) means to flip all bulbs xxx such that L≤x≤RL \leq x \leq RL≤x≤R. So for example, FLIP(3,5)FLIP(3, 5)FLIP(3,5) means to flip bulbs 333 , 444 and 555, and FLIP(5,5)FLIP(5, 5)FLIP(5,5) means to flip bulb 555.Given the value of NNN and a sequence of MMM flips, count the number of light bulbs that will be on at the end state.

输入：

The first line of the input gives the number of test cases, TTT. TTT test cases follow. Each test case starts with a line containing two integers NNN and MMM, the number of light bulbs and the number of operations, respectively. Then, there are MMM more lines, the iii-th of which contains the two integers LiL_iLi and RiR_iRi, indicating that the iii-th operation would like to flip all the bulbs from LiL_iLi to RiR_iRi , inclusive.1≤T≤10001 \leq T \leq 10001≤T≤10001≤N≤1061 \leq N \leq 10^61≤N≤1061≤M≤10001 \leq M \leq 10001≤M≤10000≤Li≤Ri≤N−10 \leq L_i \leq R_i \leq N-10≤Li≤Ri≤N−1

输出：

For each test case, output one line containing Case #x: y, where xxx is the test case number (starting from 111) and yyy is the number of light bulbs that will be on at the end state, as described above.

样例输入：

样例输出：

Case #1: 4
Case #2: 3

code：

这道题，当时写的时候，实在是一言难尽，一开始感觉是线段树？？然后写了写发现超内存，然后换成树状数组？？然后写了写发现超时，超时在分个块查询？？？然后怎么分都是超时，超时，超时，，，，自闭。。。。。
代码其实很简单，初学者都能看懂，只是这个思路，有点难想

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn=1e6+20;
int num[maxn*2];
int main()
{
 int ttt;
 scanf("%d",&ttt);
 for(int kk=1;kk<=ttt;kk++)
 {
  int n,m;
  scanf("%d%d",&n,&m);
  int count=0;
  for(int i=0;i<m;i++)
  {
   int xx,yy;
   scanf("%d%d",&xx,&yy);
   num[count++]=xx;
   num[count++]=yy+1;
  }
  sort(num,num+m*2);
  int sum=0;
  for(int i=0;i<count;i+=2)
  {
   sum+=num[i+1]-num[i];
  }
  printf("Case #%d: %d\n",kk,sum);
 }
 return 0;
 }