HDU 1166 - 敌兵布阵 (线段树,梳妆数组)


http://acm.hdu.edu.cn/showproblem.php?pid=1166
Problem Description
C国的死对头A国这段时间正在进行军事演习,所以C国间谍头子Derek和他手下Tidy又开始忙乎了。A国在海岸线沿直线布置了N个工兵营地,Derek和Tidy的任务就是要监视这些工兵营地的活动情况。由于采取了某种先进的监测手段,所以每个工兵营地的人数C国都掌握的一清二楚,每个工兵营地的人数都有可能发生变动,可能增加或减少若干人手,但这些都逃不过C国的监视。
中央情报局要研究敌人究竟演习什么战术,所以Tidy要随时向Derek汇报某一段连续的工兵营地一共有多少人,例如Derek问:“Tidy,马上汇报第3个营地到第10个营地共有多少人!”Tidy就要马上开始计算这一段的总人数并汇报。但敌兵营地的人数经常变动,而Derek每次询问的段都不一样,所以Tidy不得不每次都一个一个营地的去数,很快就精疲力尽了,Derek对Tidy的计算速度越来越不满:"你个死肥仔,算得这么慢,我炒你鱿鱼!”Tidy想:“你自己来算算看,这可真是一项累人的工作!我恨不得你炒我鱿鱼呢!”无奈之下,Tidy只好打电话向计算机专家Windbreaker求救,Windbreaker说:“死肥仔,叫你平时做多点acm题和看多点算法书,现在尝到苦果了吧!”Tidy说:"我知错了。。。"但Windbreaker已经挂掉电话了。Tidy很苦恼,这么算他真的会崩溃的,聪明的读者,你能写个程序帮他完成这项工作吗?不过如果你的程序效率不够高的话,Tidy还是会受到Derek的责骂的.
Input
第一行一个整数T,表示有T组数据。
每组数据第一行一个正整数N(N<=50000),表示敌人有N个工兵营地,接下来有N个正整数,第i个正整数ai代表第i个工兵营地里开始时有ai个人(1<=ai<=50)。
接下来每行有一条命令,命令有4种形式:
(1) Add i j,i和j为正整数,表示第i个营地增加j个人(j不超过30)
(2)Sub i j ,i和j为正整数,表示第i个营地减少j个人(j不超过30);
(3)Query i j ,i和j为正整数,i<=j,表示询问第i到第j个营地的总人数;
(4)End 表示结束,这条命令在每组数据最后出现;
每组数据最多有40000条命令
Output
对第i组数据,首先输出“Case i:”和回车,
对于每个Query询问,输出一个整数并回车,表示询问的段中的总人数,这个数保持在int以内。
Sample Input
1 10 1 2 3 4 5 6 7 8 9 10 Query 1 3 Add 3 6 Query 2 7 Sub 10 2 Add 6 3 Query 3 10 End
Sample Output
Case 1: 6 33 59

http://blog.csdn.net/liang5630/article/details/12357775?utm_source=tuicool
  1. //线段树版:  
  2. using namespace std;  
  3. int n,s[50005*4];  
  4. void update(int id,int d,int L,int R,int O)  
  5. {  
  6.     int M=L+(R-L)/2;  
  7.     if(L==R) s[O]+=d;  
  8.     else  
  9.     {  
  10.         if(id<=M) update(id,d,L,M,O*2); else update(id,d,M+1,R,O*2+1);  
  11.         s[O]=s[O*2]+s[O*2+1];  
  12.     }  
  13. }  
  14. int query(int ql,int qr,int L,int R,int O)  
  15. {  
  16.     int M=L+(R-L)/2,ans=0;  
  17.     if(ql<=L&&R<=qr) return s[O];  
  18.     if(ql<=M) ans+=query(ql,qr,L,M,O*2);  
  19.     if(M<qr)  ans+=query(ql,qr,M+1,R,O*2+1);  
  20.     return ans;  
  21. }  

  1. //树状数组版:  
  2. using namespace std;  
  3. int n,s[50005];  
  4. int LowBit(int x)  
  5. {    return x&-x;  }  
  6. void Add(int x,int d)  
  7. {  
  8.     while(x<=n)  
  9.     {  s[x]+=d;  x+=LowBit(x);}  
  10. }  
  11. int Sum(int x)  
  12. {  
  13.     int ans=0;  
  14.     while(x>0)  
  15.     {    ans+=s[x];  x-=LowBit(x); }  
  16.     return ans;  
  17. }  

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