Segment Game

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 446    Accepted Submission(s): 95

Problem Description

Lillian is a clever girl so that she has lots of fans and often receives gifts from her fans.

One day Lillian gets some segments from her fans Lawson with lengths of 1,2,3... and she intends to display them by adding them to a number line.At the i-th add operation,she will put the segment with length of i on the number line.Every time she put the segment on the line,she will count how many entire segments on that segment.During the operation ,she may delete some segments on the line.(Segments are mutually independent)

 

 

Input

There are multiple test cases.

The first line of each case contains a integer n — the number of operations(1<=n<=

2∗10⁵,∑n<=7∗10⁵)

Next n lines contain the descriptions of the operatons,one operation per line.Each operation contains two integers a , b. 

if a is 0,it means add operation that Lilian put a segment on the position b(|b|<10⁹) of the line.

(For the i-th add operation,she will put the segment on [b,b+i] of the line, with length of i.)

if a is 1,it means delete operation that Lilian will delete the segment which was added at the b-th add operation.

 

 

Output

For i-th case,the first line output the test case number.

Then for each add operation,ouput how many entire segments on the segment which Lillian newly adds.

 

 

Sample Input

3

0 0

0 3

0 1

5

0 1

0 0

1 1

0 1

0 0

 

 

Sample Output

Case #1:

0

0

0

Case #2:

0

1

0

2

 

Hint

For the second case in the sample: At the first add operation,Lillian adds a segment [1,2] on the line. At the second add operation,Lillian adds a segment [0,2] on the line. At the delete operation,Lillian deletes a segment which added at the first add operation. At the third add operation,Lillian adds a segment [1,4] on the line. At the fourth add operation,Lillian adds a segment [0,4] on the line

 

 

Source

 1004

 

解题：树状数组，神奇。。。哎。。这么简单怎么当场没想到

 

1 #include 
       
         2 using namespace std; 3 const int maxn = 400010; 4 int op[maxn],L[maxn],R[maxn],n; 5 int C[2][maxn],Li[maxn]; 6 inline int lowbit(int x) { 7     return x&(-x); 8 } 9 void update(int i,int val,int o) {10     for(; i < maxn; i += lowbit(i))11         C[o][i] += val;12 }13 int calc(int i,int o) {14     int sum = 0;15     for(; i; i -= lowbit(i))16         sum += C[o][i];17     return sum;18 }19 int main() {20     int add,tot,cnt,cs = 1;21     while(~scanf("%d",&n)) {22         memset(C,0,sizeof C);23         for(int i = add = tot = cnt = 0; i < n; ++i) {24             scanf("%d%d",op+i,L+i);25             if(op[i] == 0) {26                 R[i] = L[i] + (++add);27                 Li[tot++] = L[i];28                 Li[tot++] = R[i];29             }30         }31         sort(Li,Li + tot);32         tot = unique(Li,Li + tot) - Li;33         printf("Case #%d:\n",cs++);34         for(int i = 0; i < n; ++i) {35             if(!op[i]) {36                 int LL = lower_bound(Li,Li + tot,L[i]) - Li + 1;37                 int RR = lower_bound(Li,Li + tot,R[i]) - Li + 1;38                 printf("%d\n",calc(RR,1) - calc(LL-1,0));39                 L[cnt] = LL;40                 R[cnt++] = RR;41                 update(LL,1,0);42                 update(RR,1,1);43             } else {44                 update(L[L[i]-1],-1,0);45                 update(R[L[i]-1],-1,1);46             }47         }48     }49     return 0;50 }