HDU 1394 Minimum Inversion Number(线段树,单节点修改,区间求和)
Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21314 Accepted Submission(s): 12780
Problem 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.
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
Author
CHEN, Gaoli
类似树状数组, 线段树 建立 ,把 数的叶子作为每个数对应的位置, 然后根节点表示数目, 最后枚举所有的点 求逆序数, (单节点增加 , 区间求和)
逆序数 表达为 n-x-x-1;
代码:
#include <iostream> #include <stdio.h> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <queue> #include <stack> #include <stdlib.h> #include <list> #include <map> #include <set> #include <bitset> #include <vector> #define mem(a,b) memset(a,b,sizeof(a)) #define findx(x) lower_bound(b+1,b+1+bn,x)-b #define FIN freopen("input.txt","r",stdin) #define FOUT freopen("output.txt","w",stdout) #define S1(n) scanf("%d",&n) #define S2(n,m) scanf("%d%d",&n,&m) #define Pr(n) printf("%d\n",n) using namespace std; typedef long long ll; const double PI=acos(-1); const int INF=0x3f3f3f3f; const double esp=1e-6; const int maxn=1e6+5; const int MOD=1e9+7; const int mod=1e9+7; int dir[5][2]={0,1,0,-1,1,0,-1,0}; ll inv[maxn*2],fac[maxn]; ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;} ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;} ll lcm(ll a,ll b){ return b/gcd(a,b)*a;} ll qpow(ll x,ll n){ll res=1;for(;n;n>>=1){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;}return res;} void INV(){inv[1] = 1;for(int i = 2; i < maxn; i++) inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;} void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){ x=1; y=0; d=a; }else{ ex_gcd(b,a%b,d,y,x); y-=x*(a/b);}} void Fac(){fac[0]=1;for(int i=1;i<=maxn;i++)fac[i]=(fac[i-1]*i)%MOD;} ll inv_exgcd(ll a,ll n){ll d,x,y;ex_gcd(a,n,d,x,y);return d==1?(x+n)%n:-1;} ll inv1(ll b){return b==1?1:(MOD-MOD/b)*inv1(MOD%b)%MOD;} ll inv2(ll b){return qpow(b,MOD-2);} ll cal(ll m,ll n){if(m<n)return 0;return (fac[m]*inv[fac[n]]%MOD)%MOD*inv[fac[m-n]]%MOD;} ll cals(ll m,ll n){if(m<n)return 0;return (fac[m]*inv1(fac[n])%MOD)%MOD*inv1(fac[m-n])%MOD;} int n; struct SegTree{ int val; int addmark; }Seg[maxn]; void PushUP(int root) { Seg[root].val=Seg[root*2+1].val+Seg[root*2+2].val; } void build(int root,int begin,int end) { Seg[root].addmark=0; Seg[root].val=0; if(begin==end) return ; //scanf("%d",&Seg[root].val); else { int mid=(begin+end)>>1; build(root*2+1,begin,mid); build(root*2+2,mid+1,end); PushUP(root); } } int query(int root,int begin,int end,int left,int right) { if(left<=begin&&right>=end) return Seg[root].val; int mid=(begin+end)>>1; int ans=0; if(left<=mid) ans+= query(root*2+1,begin,mid,left,right); if(right>mid) ans+= query(root*2+2,mid+1,end,left,right); //cout<<"ans="<<ans<<endl; return ans; } void insert(int root,int l,int r,int idex,int add) { if(l==r) { Seg[root].val+=add; return; } int mid=(l+r)>>1; if(idex<=mid) insert(root*2+1,l,mid,idex,add); else insert(root*2+2,mid+1,r,idex,add); PushUP(root); } int x[maxn]; int main() { while(~S1(n)) { mem(x,0); build(1,0,n-1); int sum=0; for(int i=0;i<n;i++) { S1(x[i]); sum+=query(1,0,n-1,x[i],n-1); insert(1,0,n-1,x[i],1); } int ans=sum; //cout<<"sum==="<<ans<<endl; for(int i=0;i<n;i++) { sum+=n-x[i]-x[i]-1; ans=min(ans,sum); } Pr(ans); } return 0; }
123
岂曰无衣?与子同袍。王于兴师,修我戈矛。与子同仇!
岂曰无衣?与子同泽。王于兴师,修我矛戟。与子偕作!
岂曰无衣?与子同裳。王于兴师,修我甲兵。与子偕行!