HDU 1394 Minimum Inversion Number(归并排序 线段树 逆序数)
http://acm.hdu.edu.cn/showproblem.php?pid=1394
Description
The inversion number of a given number sequence
a1, a2, ..., an is the number of pairs (ai, aj)
that satisfy i < j and ai > aj.
For a given sequence of numbers a1, a2, ..., an,
if we move the first m >= 0 numbers to the end of the seqence,
we will obtain another sequence.
There are totally n such sequences as the following:
a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)
You are asked to write a program
to find the minimum inversion number out of the above sequences.
Input
The input consists of a number of test cases.
Each case consists of two lines:
the first line contains a positive integer n (n <= 5000);
the next line contains a permutation of the n integers from 0 to n-1.
Output
For each case,
output the minimum inversion number on a single line.
Sample Input
10
1 3 6 9 0 8 5 7 4 2
Sample Output
16
题意:
有n个从0到n 的数 按照给定次序排列 每次只能将最前的一个数移到末尾 算出每种情况的逆序数 并求出逆序数最小值
思路:
归并排序或线段树
首先 我们把当前情况下的逆序数求出
然后循环n次 每次最前一个往最后移 则 ans=ans+ (n-1+a[i])-a[i]
#include<stdio.h> #include<string.h> #include<math.h> #include<iostream> #include<algorithm> using namespace std; int num[5000+10],a[5000+10],t[5000+10]; //int vis[5000+10]; int ans; void Merge(int x,int m,int y) { int p=x,q=m+1,i=x; while(p<=m||q<=y) { if(q>y||(p<=m&&a[p]<=a[q])) t[i++]=a[p++]; else { t[i++]=a[q++]; ans+=m-p+1; } } for(i=x;i<=y;i++) a[i]=t[i]; } void mergesort(int x,int y) { if(x!=y) { int mid=(x+y)/2; mergesort(x,mid); mergesort(mid+1,y); Merge(x,mid,y); } } int main() { int n,i,j,minn; while(scanf("%d",&n)!=EOF) { ans=0; for(i=1;i<=n;i++) {scanf("%d",&a[i]); num[i]=a[i];} mergesort(1,n); minn=ans; for(i=1;i<n;i++) { ans=ans+n-1-2*num[i]; if(ans<minn) minn=ans; } cout<<minn<<endl; } return 0; }
线段树的版本
对于初始序列
每次加入一个数就对其标记 并且查询是否有比他大的数已经被加入 从而求出逆序数
#include<stdio.h> #include<string.h> #include<math.h> #include<iostream> #include<algorithm> #include<queue> #include<stack> #define mem(a,b) memset(a,b,sizeof(a)) #define ll __int64 #define MAXN 1000 #define INF 0x7ffffff #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 using namespace std; int sum[50000]; int a[5000+10]; void pushup(int rt) { sum[rt]=sum[rt<<1]+sum[rt<<1|1]; } void build(int l,int r,int rt) { if(l==r) { sum[rt]=0; return ; } int m=(l+r)>>1; build(lson); build(rson); pushup(rt); } int getnum(int L,int R,int l,int r,int rt) { if(L<=l&&r<=R) { return sum[rt]; } int ret=0; int m=(l+r)>>1; if(L<=m) ret+=getnum(L,R,lson); if(R>m) ret+=getnum(L,R,rson); return ret; } void update(int num,int l,int r,int rt) { if(l==r&&l==num) { sum[rt]=1; return ; } int m=(l+r)>>1; if(num<=m) update(num,lson); else update(num,rson); pushup(rt); } int main() { int n; int i,j; int ans,minn; while(scanf("%d",&n)!=EOF) { ans=0; build(0,n-1,1); for(i=0;i<n;i++) { scanf("%d",&a[i]); ans+=getnum(a[i],n-1,0,n-1,1); update(a[i],0,n-1,1); } minn=ans; for(i=0;i<n-1;i++) { ans=ans-a[i]*2+n-1; if(minn>ans) minn=ans; } cout<<minn<<endl; } return 0; }