题目:https://vjudge.net/problem/HDU-1394
题目大意
给定一个序列,将前m(0-n-1)个数字移到末尾,得到一个新的序列。求所有这些序列的逆序的最小值
分析
如果知道了当前序列逆序数为sum,那么移动头元素后的逆序数将会是sum-x+(n-1-x), 那么就是怎么求当前逆序数,自然就用到了线段树。
代码
/********************************************************************
* File Name: hdu1394.cpp
* Author: Sequin
* mail: Catherine199787@outlook.com
* Created Time: 三 9/27 11:29:02 2017
*************************************************************************/
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <ctype.h>
#include <set>
#include <vector>
#include <cmath>
#include <bitset>
#include <algorithm>
#include <climits>
#include <string>
#include <list>
#include <cctype>
#include <cstdlib>
#include <fstream>
#include <sstream>
using namespace std;
#define lson node << 1
#define rson node << 1 | 1
#define LS l,mid,lson
#define RS mid+1,r,rson
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define pi acos(-1.0)
#define pii pair<int,int>
#define ll long long
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
#define maxn 5005
int segTree[maxn << 2];
int num[maxn];
void build(int l, int r, int node) {
if(l == r) {
segTree[node] = 0;
}
else{
int mid = (l + r) >> 1;
build(LS);
build(RS);
}
}
int query(int a, int b, int l, int r, int node) {
if(l >= a && r <= b) {
return segTree[node];
}
int re = 0;
int mid = (l + r) >> 1;
if(a <= mid) {
re += query(a, b, LS);
}
if(mid < b) {
re += query(a, b, RS);
}
return re;
}
void update(int index, int l, int r, int node) {
if(l == r) {
segTree[node]++;
return;
}
int mid = (l + r) >> 1;
if(index <= mid) {
update(index, LS);
}
else{
update(index, RS);
}
segTree[node] = segTree[lson] + segTree[rson];
}
int main() {
int n;
while(cin >> n) {
MEM(segTree, 0);
build(0, n-1, 1);
for(int i = 0; i < n; i++) {
cin >> num[i];
}
int ret = 0;
for(int i = 0; i < n; i++) {
ret += query(num[i], n-1, 0, n-1, 1);
update(num[i], 0, n-1, 1);
}
int ans = ret;
for(int i = 0; i < n; i++) {
ret = ret - num[i] + n - num[i] - 1;
ans = min(ret, ans);
}
cout << ans << endl;
}
return 0;
}