poj2299--B - Ultra-QuickSort(线段树,离散化)
Ultra-QuickSort
Time Limit: 7000MS | Memory Limit: 65536K | |
Total Submissions: 41215 | Accepted: 14915 |
Description
![](http://poj.org/images/2299_1.jpg)
Ultra-QuickSort produces the output
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
5 9 1 0 5 4 3 1 2 3 0
Sample Output
6 0
题意:求出最小交换次数。使得数组变得有序(由小到大)。 也就是求出逆序数。这个题归并排序,树状数组。线段树又能够解,能够当做练手用
使用线段树解的话,由于给出的数值非常大。所以须要离散化。一開始直接使用map果断的超时了,仅仅能想一个更简便的离散的方法,使用结构体存下一開始的值k和它初始的序号(id1),sort对k进行排序。得到新的序号(id2),通过id1直接改变给出的数组。变为id2。这样仅仅用n + logn的时间就能够离散完毕,(注意要推断反复的值。反复的值共享一个id2)。至于线段树部分就是一个模板
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define LL __int64 #define maxn 600000 #define lmin 1 #define rmax n #define lson l,(l+r)/2,rt<<1 #define rson (l+r)/2+1,r,rt<<1|1 #define root lmin,rmax,1 #define now l,r,rt #define int_now LL l,LL r,LL rt struct node{ LL id1 , id2 ; LL k ; }p[maxn] ; LL cl[maxn<<2] , a[maxn] ; bool cmp(node a,node b) { return a.k < b.k ; } void push_up(int_now) { cl[rt] = cl[rt<<1] + cl[rt<<1|1] ; } void update(LL i,int_now) { if( i < l || i > r ) return ; if( i == l && i==r ) { cl[rt]++ ; return ; } update(i,lson); update(i,rson); push_up(now); return ; } LL query(int ll,int rr,int_now) { if( ll > r || rr < l ) return 0; if( ll <= l && r <= rr ) return cl[rt] ; return query(ll,rr,lson) + query(ll,rr,rson); } int main() { LL i , n , m , l , r , x , num ; while(scanf("%I64d", &m) && m) { for(i = 0 ; i < m ; i++) { scanf("%I64d", &a[i]); p[i].k = a[i] ; p[i].id1 = i ; } sort(p,p+m,cmp); int temp = -1 ; n = 0 ; for(i = 0 ; i < m ; i++) { if( p[i].k == temp ) p[i].id2 = n ; else { p[i].id2 = ++n ; temp = p[i].k ; } } for(i = 0 ; i < m ; i++) a[ p[i].id1 ] = p[i].id2 ; memset(cl,0,sizeof(cl)); num = 0 ; for(i = 0 ; i < m ; i++) { num += (i - query(1,a[i],root)); update(a[i],root); } printf("%I64d\n", num); } return 0; }