hdu3874-Necklace 线段树+离散化

Necklace

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4849    Accepted Submission(s): 1659

Problem Description

Mery has a beautiful necklace. The necklace is made up of N magic balls. Each ball has a beautiful value. The balls with the same beautiful value look the same, so if two or more balls have the same beautiful value, we just count it once. We define the beautiful value of some interval [x,y] as F(x,y). F(x,y) is calculated as the sum of the beautiful value from the xth ball to the yth ball and the same value is ONLY COUNTED ONCE. For example, if the necklace is 1 1 1 2 3 1, we have F(1,3)=1, F(2,4)=3, F(2,6)=6.

Now Mery thinks the necklace is too long. She plans to take some continuous part of the necklace to build a new one. She wants to know each of the beautiful value of M continuous parts of the necklace. She will give you M intervals [L,R] (1<=L<=R<=N) and you must tell her F(L,R) of them.

 

Input

The first line is T(T<=10), representing the number of test cases.
  For each case, the first line is a number N,1 <=N <=50000, indicating the number of the magic balls. The second line contains N non-negative integer numbers not greater 1000000, representing the beautiful value of the N balls. The third line has a number M, 1 <=M <=200000, meaning the nunber of the queries. Each of the next M lines contains L and R, the query.

 

Output

For each query, output a line contains an integer number, representing the result of the query.

 

Sample Input

  

   2
6
1 2 3 4 3 5
3
1 2
3 5
2 6
6
1 1 1 2 3 5
3
1 1
2 4
3 5
  

 

Sample Output

  

   3
7
14
1
3
6
  
  


  
  


  
  

   题意：有一串项链，项链上的每一颗珍珠都有一定的价值，看起来一样的珍珠的价值是一样的，现在需要求从第i颗到第j颗珍珠的价值和，注意的是如果有相同的珍珠则只计算一次。
  
  


  
  

   解题思路：如果按照普通的思路这道题会超时，在这里我们需要用到的是线段树求区间和以及离散化缩小数据规模，离散化的具体知识可以看我的博客：
   点击打开链接http://blog.youkuaiyun.com/wang_heng199/article/details/75324299
  
  


  
  

   ac代码：
  
  

   
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define lson l,m,o<<1
#define rson m+1,r,o<<1|1
typedef __int64 ll;
int a[50010];
int n;
ll sum[50010<<2];
struct node{
    int l;
    int r;
    int i;
};

node q[200010];
int vis[1000010];
ll ans[200010];

void pushup(int o)
{
    sum[o]=sum[o<<1]+sum[o<<1|1];
}
void build()
{
    memset(sum,0,sizeof(sum));
}

void update(int l,int r,int o,int p,int x)
{
    if(l==r)
    {
        sum[o]+=x;
        return;
    }
    int m=l+r>>1;
    if(p<=m)update(lson,p,x);
    else update(rson,p,x);
    pushup(o);
}

ll query(int l,int r,int o,int L,int R)
{
    if(L<=l&&r<=R)return sum[o];

    int m=l+r>>1;
    ll ans=0;
    if(L<=m)ans+=query(lson,L,R);
    if(m<R)ans+=query(rson,L,R);
    return ans;
}

bool cmp(node a,node b)
{
    return a.r<b.r;
}

int main()
{
    int t;
    scanf("%d",&t);

    while(t--)
    {
        int i;

        scanf("%d",&n);
        for(i=0;i<n;i++)scanf("%d",&a[i]);

        build();


        int Q;

        scanf("%d",&Q);

        for(i=0;i<Q;i++){
        scanf("%d%d",&q[i].l,&q[i].r);
        q[i].i=i;
        }

        sort(q,q+Q,cmp);
        int ll=0;
        memset(vis,-1,sizeof(vis));
        for(i=0;i<Q;i++)
        {
            while(ll<=q[i].r-1)
            {
                if(vis[a[ll]]!=-1)update(0,n-1,1,vis[a[ll]],-a[ll]);

                vis[a[ll]]=ll;

                update(0,n-1,1,ll,a[ll]);
                ll++;
            }
            ans[q[i].i]=query(0,n-1,1,q[i].l-1,q[i].r-1);
        }

        for(i=0;i<Q;i++)printf("%I64d\n",ans[i]);
    }

    return 0;
}
   

    

   
题目链接：
   点击打开链接http://acm.hdu.edu.cn/showproblem.php?pid=3874
   

   

  
  

   推荐相似类型题目：
   点击打开链接http://blog.youkuaiyun.com/wang_heng199/article/details/75329999