Light bulbs------The Preliminary Contest for ICPC Asia Shanghai 2019
There are NN light bulbs indexed from 00 to N-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) means to flip all bulbs xxsuch that L \leq x \leq RL≤x≤R. So for example, FLIP(3, 5)FLIP(3,5) means to flip bulbs 33 , 44 and 55, and FLIP(5, 5)FLIP(5,5) means to flip bulb 55.
Given the value of NN and a sequence of MM flips, count the number of light bulbs that will be on at the end state.
InputFile
The first line of the input gives the number of test cases, TT. TT test cases follow. Each test case starts with a line containing two integers NN and MM, the number of light bulbs and the number of operations, respectively. Then, there are MM more lines, the ii-th of which contains the two integers L_iLi and R_iRi, indicating that the ii-th operation would like to flip all the bulbs from L_iLi to R_iRi , inclusive.
1 \leq T \leq 10001≤T≤1000
1 \leq N \leq 10^61≤N≤106
1 \leq M \leq 10001≤M≤1000
0 \leq L_i \leq R_i \leq N-10≤Li≤Ri≤N−1
OutputFile
For each test case, output one line containing Case #x: y, where xx is the test case number (starting from 11) and yy is the number of light bulbs that will be on at the end state, as described above.
样例输入
2
10 2 2 6 4 8 6 3 1 1 2 3 3 4
样例输出
Case #1: 4
Case #2: 3
初始代码：
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 6 struct node 7 { 8 int data; 9 bool flag; 10 }str[20006]; 11 12 bool cmp(struct node a, struct node b) 13 { 14 if(a.data!=b.data) return a.data <= b.data; 15 else return a.flag==true; 16 } 17 18 int main() 19 { 20 int t, m, n, i, tot, top, x, pro, loc, re; 21 scanf("%d", &t); 22 tot = 0; 23 while(t--) 24 { 25 scanf("%d %d", &n, &m); 26 top = 0; 27 re = 0; 28 for(i=0;i<m;i++) 29 { 30 scanf("%d %d", &str[top].data, &str[top+1].data); 31 str[top+1].data++; 32 str[top].flag = true; 33 str[top+1].flag = false; 34 top+=2; 35 } 36 37 sort(str, str+top, cmp); 38 39 x = 0; 40 pro = str[0].data; 41 42 loc = 0; 43 while(loc<top) 44 { 45 if(str[loc].flag==true) 46 { 47 x++; 48 if(x%2==1) re = re + str[loc].data - pro; 49 pro = str[loc].data; 50 loc++; 51 while(str[loc].data==pro) 52 { 53 if(str[loc].flag==true) x++; 54 else x--; 55 pro = str[loc].data; 56 loc++; 57 } 58 59 if(x%2==1) re = re + str[loc].data - pro; 60 pro = str[loc].data; 61 } 62 63 else 64 { 65 x--; 66 if(x%2==1) re = re + str[loc].data - pro; 67 pro = str[loc].data; 68 loc++; 69 while(str[loc].data==pro) 70 { 71 if(str[loc].flag==true) x++; 72 else x--; 73 pro = str[loc].data; 74 loc++; 75 } 76 77 if(x%2==1) re = re + str[loc].data - pro; 78 pro = str[loc].data; 79 } 80 81 } 82 tot++; 83 printf("Case #%d: %d\n", tot, re); 84 } 85 return 0; 86 }
简洁版代码：
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 int main() 6 { 7 int t, m, n, i, tot, top, x, pro, loc, re; 8 int a[2006]; 9 scanf("%d", &t); 10 tot = 0; 11 while(t--) 12 { 13 scanf("%d %d", &n, &m); 14 top = 0; 15 re = 0; 16 for(i=0;i<m;i++) 17 { 18 scanf("%d %d", &a[top], &a[top+1]); 19 a[top+1]++; 20 top+=2; 21 } 22 23 sort(a, a+top); 24 25 x = 0; 26 pro = a[0]; 27 28 loc = 0; 29 while(loc<top) 30 { 31 x++; 32 if(x%2==1) re = re + a[loc] - pro; 33 pro = a[loc]; 34 loc++; 35 while(a[loc]==pro) 36 { 37 x++; 38 pro = a[loc]; 39 loc++; 40 } 41 42 if(x%2==1) re = re + a[loc] - pro; 43 pro = a[loc]; 44 } 45 tot++; 46 printf("Case #%d: %d\n", tot, re); 47 } 48 return 0; 49 }
