POJ 2299 离散化+树状数组

这段时间没怎么发博客 被动态规划折磨的头疼...

今天就把dp放下了 开始上手了几道树状数组的题

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 60829 Accepted: 22563

Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence 
9 1 0 5 4 ,
Ultra-QuickSort produces the output 
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

59105431230

Sample Output

60

Source

Waterloo local 2005.02.05

#include<algorithm>#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<string>#include<stack>#include<queue>#include<cmath>#include<stack>#include<list>#include<map>#include<set>typedef long long ll;using namespace std;int a[500005];int bit[500005];int n;struct node{int i1;int x;}nd[500005];bool cmp1(node a,node b){    return a.x<b.x;  //根据值的大小排序}bool cmp2(node a,node b){    return a.i1<b.i1;  //之前根据值的大小排序 所以下标被大打乱了 离散化后根据下标重新排好}int sum(int i){    int s=0;    while(i>0)    {        s+=bit[i];        i-=i&-i;    }    return s;}void add(int i,int x){    while(i<=500003)    {        bit[i]+=x;        i+=i&-i;    }}int main(){    int i,j,k;    while(scanf("%d",&n)==1&&n)    {        for(i=0;i<n;i++)        {            nd[i].i1=i;            scanf("%d",&k);            nd[i].x=k;        }        memset(bit,0,sizeof bit); //不要忘记初始化!!        sort(nd,nd+n,cmp1);//根据值的大小排序        for(i=0;i<n;i++)        {            nd[i].x=i+1;  //进行离散化 因为sum函数里条件为i>0 所以最好还是不要出现0值 所以+1就好了        }        sort(nd,nd+n,cmp2); //根据下标重新排好        ll ans=0;        for(i=0;i<n;i++)        {            ans+=i-sum(nd[i].x);  //之后这就是白书上的模板了 冒泡模板            add(nd[i].x,1);        }        printf("%lld\n",ans); // ans一定要开long long型的!int型过不了!    }    return 0;}