Codeforces Gym 100114 H. Milestones 离线树状数组
H. Milestones
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100114
Description
The longest road of the Fairy Kingdom has n milestones. A long-established tradition defines a specific color for milestones in each region, with a total of m colors in the kingdom. There is a map describing all milestones and their colors. A number of painter teams are responsible for milestone maintenance and painting. Typically, each team is assigned a road section spanning from milestone #l to milestone #r. When optimizing the assignments, the supervisor often has to determine how many different colors it will take to paint all milestones in the section l…r. Example. Suppose there are five milestones #1, #2, #3, #4, #5 to be painted with colors 1, 2, 3, 2, 1, respectively. In this case, only two different paints are necessary for milestones 2…4: color 2 for milestones #2 and #4, and color 3 for milestone #3. Write a program that, given a map, will be able to handle multiple requests of the kind described above.
Input
Output
Sample Input
5 3 1 2 3 2 1 1 5 1 3 2 4
Sample Output
HINT
题意
求区间内有多少个不同的数
没有修改
题解:
离线维护树状数组就好了
代码:
#include <cstdio>
#include <cstdlib>
#include <sstream>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <utility>
#include <vector>
#include <queue>
#include <map>
#include <set>
using namespace std; typedef long long ll;
typedef pair<int,int> PII;
#define DEBUG(x) cout<< #x << ':' << x << endl
#define FOR(i,s,t) for(int i = (s);i <= (t);i++)
#define FORD(i,s,t) for(int i = (s);i >= (t);i--)
#define REP(i,n) for(int i=0;i<(n);i++)
#define REPD(i,n) for(int i=(n-1);i>=0;i--)
#define PII pair<int,int>
#define PB push_back
#define ft first
#define sd second
#define lowbit(x) (x&(-x))
#define INF (1<<30)
#define eps (1e-8) const int maxq = ;
const int maxn = ;
int a[maxn],C[maxn],last[];
int ans[maxq];
void init(){
memset(C,,sizeof(C));
memset(last,-,sizeof(last));
}
struct Query{
int l,r;
int idx;
bool operator < (const Query & rhs)const{
return r < rhs.r;
}
}Q[maxq]; void add(int x,int val){
while(x<maxn){
C[x] += val;
x += lowbit(x);
}
}
int sum(int x){
int res = ;
while(x > ){
res += C[x];
x -= lowbit(x);
}
return res;
}
int main(){
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
int n;
while(~scanf("%d",&n)){
init();
int q;
scanf("%d",&q);
FOR(i,,n)scanf("%d",&a[i]);
REP(i,q){
scanf("%d%d",&Q[i].l,&Q[i].r);
Q[i].idx = i;
}
sort(Q,Q+q);
int pre = ;
REP(i,q){
FOR(j,pre,Q[i].r){
if(last[a[j]]==-){
add(j,);
}else {
add(last[a[j]],-);
add(j,);
}
last[a[j]] = j;
}
ans[Q[i].idx] = sum(Q[i].r)-sum(Q[i].l-);
pre = Q[i].r+;
}
REP(i,q)printf("%d\n",ans[i]);
}
return ;
}
Codeforces Gym 100114 H. Milestones 离线树状数组的更多相关文章
- Codeforces Gym 100269F Flight Boarding Optimization 树状数组维护dp
Flight Boarding Optimization 题目连接: http://codeforces.com/gym/100269/attachments Description Peter is ...
- Educational Codeforces Round 10 D. Nested Segments 离线树状数组 离散化
D. Nested Segments 题目连接: http://www.codeforces.com/contest/652/problem/D Description You are given n ...
- Codeforces Round #365 (Div. 2) D - Mishka and Interesting sum(离线树状数组)
http://codeforces.com/contest/703/problem/D 题意: 给出一行数,有m次查询,每次查询输出区间内出现次数为偶数次的数字的异或和. 思路: 这儿利用一下异或和的 ...
- CodeForces - 220B Little Elephant and Array (莫队+离散化 / 离线树状数组)
题意:N个数,M个查询,求[Li,Ri]区间内出现次数等于其数值大小的数的个数. 分析:用莫队处理离线问题是一种解决方案.但ai的范围可达到1e9,所以需要离散化预处理.每次区间向外扩的更新的过程中, ...
- CF #365 (Div. 2) D - Mishka and Interesting sum 离线树状数组
题目链接:CF #365 (Div. 2) D - Mishka and Interesting sum 题意:给出n个数和m个询问,(1 ≤ n, m ≤ 1 000 000) ,问在每个区间里所有 ...
- CF #365 (Div. 2) D - Mishka and Interesting sum 离线树状数组(转)
转载自:http://www.cnblogs.com/icode-girl/p/5744409.html 题目链接:CF #365 (Div. 2) D - Mishka and Interestin ...
- 离线树状数组 hihocoder 1391 Countries
官方题解: // 离线树状数组 hihocoder 1391 Countries #include <iostream> #include <cstdio> #include ...
- 区间的关系的计数 HDU 4638 离线+树状数组
题目大意:给你n个人,每个人都有一个id,有m个询问,每次询问一个区间[l,r],问该区间内部有多少的id是连续的(单独的也算是一个) 思路:做了那么多离线+树状数组的题目,感觉这种东西就是一个模板了 ...
- BZOJ_2743_[HEOI2012]采花_离线+树状数组
BZOJ_2743_[HEOI2012]采花_离线+树状数组 Description 萧芸斓是Z国的公主,平时的一大爱好是采花.今天天气晴朗,阳光明媚,公主清晨便去了皇宫中新建的花园采花 .花园足够大 ...
随机推荐
- Entity Framework中编辑时错误ObjectStateManager 中已存在具有同一键的对象
ObjectStateManager 中已存在具有同一键的对象.ObjectStateManager 无法跟踪具有相同键的多个对象. 说明: 执行当前 Web 请求期间,出现未经处理的异常.请检查堆栈 ...
- mybatis+spring+struts2框架整合
近期公司要开发新的项目,要用struts2+mybatis+spring框架,所以学习了下,来自己的博客发表下,希望能给大家带来帮助!下边我把我的myschool开发的源代码以及数据库贴出来! 开 ...
- centos使用网易163yum源
CentOS系统自带的更新源的速度实在是慢,为了让CentOS6使用速度更快的YUM更新源,可以选择163(网易)的更新源. 1.下载repo文件 wget http://mirrors.163.co ...
- POJ 2549 Sumsets
Sumsets Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10593 Accepted: 2890 Descript ...
- Druid连接池简单入门
偶尔的机会解释Druid连接池,后起之秀,但是评价不错,另外由于是阿里淘宝使用过的所以还是蛮看好的. 1.jar包依赖--Druid依赖代码 <dependency> <groupI ...
- 数组乘积--满足result[i] = input数组中除了input[i]之外所有数的乘积(假设不会溢出
数组乘积(15分) 输入:一个长度为n的整数数组input 输出:一个长度为n的整数数组result,满足result[i] = input数组中除了input[i]之外所有数的乘积(假设不会溢出). ...
- 【剑指offer 面试题23】从上往下打印二叉树
思路: 没啥好说的,BFS. C++: #include <iostream> #include <queue> using namespace std; struct Tre ...
- 《浅析各类DDoS攻击放大技术》
原文链接:http://www.freebuf.com/articles/network/76021.html FreeBuf曾报道过,BT种子协议家族漏洞可用作反射分布式拒绝服务攻击(DRDoS a ...
- mybatis系列-05-SqlMapConfig.xml详解
mybatis的全局配置文件SqlMapConfig.xml,配置内容如下: properties(属性) settings(全局配置参数) typeAliases(类型别名) typeHandler ...
- C#调用dll(C++(Win32))时的类型转换总结(转)
http://www.cnblogs.com/lidabo/archive/2012/06/05/2536737.html C++(Win 32) C# char** 作为输入参数转为char ...