hdu2838 Cow Sorting 树状数组

 

7 口碑商家客流量预测大赛》

Cow Sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3322    Accepted Submission(s): 1118

Problem Description

Sherlock's N (1 ≤ N ≤ 100,000) cows are lined up to be milked in the evening. Each cow has a unique "grumpiness" level in the range 1...100,000. Since grumpy cows are more likely to damage Sherlock's milking equipment, Sherlock would like to reorder the cows in line so they are lined up in increasing order of grumpiness. During this process, the places of any two cows (necessarily adjacent) can be interchanged. Since grumpy cows are harder to move, it takes Sherlock a total of X + Y units of time to exchange two cows whose grumpiness levels are X and Y.

Please help Sherlock calculate the minimal time required to reorder the cows.

 

Input

Line 1: A single integer: N
Lines 2..N + 1: Each line contains a single integer: line i + 1 describes the grumpiness of cow i.

 

Output

Line 1: A single line with the minimal time required to reorder the cows in increasing order of grumpiness.

 

Sample Input

3231

 

Sample Output

7
Hint
Input DetailsThree cows are standing in line with respective grumpiness levels 2, 3, and 1.Output Details2 3 1 : Initial order.2 1 3 : After interchanging cows with grumpiness 3 and 1 (time=1+3=4).1 2 3 : After interchanging cows with grumpiness 1 and 2 (time=2+1=3).
 

 

Source

2009 Multi-University Training Contest 3 - Host by WHU

get到新技巧就是树状数组不仅可以用来统计前面有多少个玩意儿比自己小等于，那几个玩意儿的值也是可以记录的，想象成大数组打标记一样。。。几天没敲电脑现在啥话也憋不出来。。。

/* ━━━━━┒ ┓┏┓┏┓┃μ'sic foever!! ┛┗┛┗┛┃＼○／ ┓┏┓┏┓┃ / ┛┗┛┗┛┃ノ) ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┃┃┃┃┃┃ ┻┻┻┻┻┻ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <map>#include <stack>#include <vector>using namespace std;const int maxn=1e5+10;long long save[maxn];struct unit{    int pos;    long long val;}c[maxn];int n;int lowbit(int x){    return x&(-x);}void update(int x,int num,long long val){    while(x<=maxn){        c[x].pos += num;        c[x].val += val;        x += lowbit(x);    }}long long getsum(long long x){    long long sum=0;    while(x){        sum += c[x].pos;        x -=lowbit(x);    }    return sum;}long long getVal(int x){    long long sum = 0;    while(x){        sum += c[x].val;        x -= lowbit(x);    }    return sum;}int main(){    int i,j;    long long ans;    while(~scanf("%d",&n)){        ans = 0;        for(i=1;i<=n;i++){            scanf("%lld",&save[i]);        }        for(i=n;i>=1;i--){            //cout<<getsum(save[i])<<"  "<<getVal(save[i])<<endl;            ans += getsum(save[i])*save[i] + getVal(save[i]);            update(save[i], 1, save[i]);        }        printf("%lld\n",ans);    }    return 0;}