HDU 4217 Data Structure?（线段树 or 树状数组）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4217

 

 

Problem Description

Data structure is one of the basic skills for Computer Science students, which is a particular way of storing and organizing data in a computer so that it can be used efficiently. Today let me introduce a data-structure-like problem for you.
Original, there are N numbers, namely 1, 2, 3...N. Each round, iSea find out the Ki-th smallest number and take it away, your task is reporting him the total sum of the numbers he has taken away.

 

 

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case includes two integers N, K, K indicates the round numbers. Then a line with K numbers following, indicating in i (1-based) round, iSea take away the Ki-th smallest away.

Technical Specification
1. 1 <= T <= 128
2. 1 <= K <= N <= 262 144
3. 1 <= Ki <= N - i + 1

 

 

Output

For each test case, output the case number first, then the sum.

 

 

Sample Input

2 3 2 1 1 10 3 3 9 1

 

 

Sample Output

Case 1: 3 Case 2: 14

 

 

Author

iSea@WHU

 

 

Source

首届华中区程序设计邀请赛暨第十届武汉大学程序设计大赛

 

 

 

 

题意：

n个数为 1 → n，一共有 k 次操作，每次取出第 ki 小的数。

问所有取出数字之和。

（树状数组）

代码如下：

 

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define maxn 262147
#define LL __int64
int c[maxn], a[maxn];
int n, k;
int Lowbit(int x)  // 2^k
{
    return x&(-x);
}
void update(int i, int x)//i点增量为x
{
    while(i <= n)
    {
        c[i] += x;
        i += Lowbit(i);
    }
}

int sum(int x)//区间求和 [1,x]
{
    int sum=0;
    while(x>0)
    {
        sum+=c[x];
        x-=Lowbit(x);
    }
    return sum;
}

int er_find(int kk)
{
    int L = kk, R = n;
    int tmp = kk;
    while(L <= R)
    {
        int mid = L+(R-L)/2;
        int tt = sum(mid);
        if(tt == kk)
        {
            tmp = mid;
            R = mid-1;
        }
        else if(tt < kk)
        {
            L = mid+1;
        }
        else if(tt > kk)
        {
            R = mid-1;
        }
    }
    return tmp;
}
int main()
{
    int t;
    int cas = 0;
    scanf("%d",&t);
    while(t--)
    {
        memset(c,0,sizeof(c));
        scanf("%d%d",&n,&k);
        for(int i = 1; i <= n; i++)
        {
            update(i,1);
        }
        int ki;
        LL ans = 0;
        for(int i = 1; i <= k; i++)
        {
            scanf("%d",&ki);
            int tmp = er_find(ki);

            ans += tmp;
            update(tmp, -1);//减一变为0
        }
        printf("Case %d: %I64d\n",++cas,ans);
    }
    return 0;
}

/*
99
3 2
1 1
10 3
3 9 1
10 5
1 2 3 4 5
10 4
1 2 3 4
10 5
5 4 3 2 1
10 5
1 5 7 4 2
*/

贴一发别人的线段树：http://www.cnblogs.com/xiaohongmao/archive/2012/04/29/2476452.html

 

#include <stdio.h>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn = 300000;
int tree[maxn<<2];
int temp;
void build(int l,int r,int rt)
{
    tree[rt]=r-l+1;
    if(l==r)
        return;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}

void dele(int del,int l,int r,int rt)
{
    tree[rt]--;//0 已取
    if(l==r)
    {
        temp=l;
        return;
    }
    int m=(l+r)>>1;
    if(del<=tree[rt<<1])
        dele(del,lson);
    else
    {
        del-=tree[rt<<1];
        dele(del,rson);
    }
}
int main()
{
    int t,n,k,ki,i;
    int nCase=1;
    __int64 sum;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        build(1,n,1);
        sum=0;
        for(i=0; i<k; i++)
        {
            scanf("%d",&ki);
            dele(ki,1,n,1);
            sum+=temp;
        }
        printf("Case %d: %I64d\n",nCase++,sum);
    }
    return 0;
}