Light bulbs------The Preliminary Contest for ICPC Asia Shanghai 2019

There are

A FLIP operation switches the state of a contiguous subset of bulbs. $\leq x \leq RL≤x≤R. So for example, FLIP(3, 5)FLIP(3,5) means to flip bulbs 33 , 44 and 55, and FLIP(5, 5)FLIP(5,5) means to flip bulb 55.$

Given the value of

InputFile

The first line of the input gives the number of test cases, $L_iLi and R_iRi, indicating that the ii-th operation would like to flip all the bulbs from L_iLi to R_iRi , inclusive.$

$\leq T \leq 10001≤T≤1000$

$\leq N \leq 10^61≤N≤106$

$\leq M \leq 10001≤M≤1000$

$\leq L_i \leq R_i \leq N-10≤Li≤Ri≤N−1$

OutputFile

For each test case, output one line containing Case #x: y, where

样例输入

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 }