http://poj.org/problem?id=2299
在兩個元素相同的數列裡,其中一個數列要移動到另一個數列相同元素相同的位置,那麼要移動的次數就是這個數列關於另一個數列的逆序對數(hash後)
逆序對的求法我原來的博文有 http://www.cnblogs.com/iwtwiioi/p/3523120.html
用歸併排序求逆序對,大的在前
左閉右開
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
const int N=500005, oo=~0u>>1;
int a[N], L[N], R[N];
long long cnt;
void gb(int l, int r) {
if(l<r-1) {
int m=(l+r)>>1, i, j;
gb(l, m); gb(m, r);
for(i=0; i<m-l; ++i) L[i]=a[l+i];
for(j=0; j<r-m; ++j) R[j]=a[m+j];
L[i]=R[j]=-oo; i=j=0;
while(l<r) {
if(L[i]>R[j]) {
a[l++]=L[i++];
cnt+=r-m-j;
}
else a[l++]=R[j++];
}
}
}
int main() {
int n;
while(scanf("%d", &n) && n) {
cnt=0;
rep(i, n) read(a[i]);
gb(0, n);
printf("%lld\n", cnt);
}
return 0;
}
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.
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
Source