hdu 2838 Cow Sorting（樹狀數組）

真是得跪,,,我以前做的求逆序數若爆了，連芒果大神都對我無語了。嗚嗚嗚

這題和poj那道置換數很像，但是這題要求必須相鄰的數才能交換。這就可以用樹狀數組求數x之前比x大的數量和比x大的數的和。

以前只是覺得在x之前插入，然後查詢的時候找比x大的，現在發現不對。

C1=a1

C2=a1+a2

C3=a3

C4=a1+a2+a3+a4

.....

樹狀數組只能求0~N的和。比如 3 : sum+=a3,x-=lowbit(x) x=2 sum+=a1+a2.

插入的情況:比如插入3,就是說C3+1,x+=lowbit(3) x=4.C[4]+=1。。。。直到n.

/*Problem ID:meaning:Analyzing:*/#include <iostream>#include <algorithm>#include<cstdio>#include<cmath>#include<cstdlib>#include<cstring>#include<vector>using namespace std;typedef struct even{int y1,y2,x;}even;#define clr(A,k) memset(A,k,sizeof(A))#define FOR(i,s,t) for(int i=(s); i<(t); i++)#define LL long long#define BUG puts("here!!!")#define print(x) printf("%d\n",x)#define STOP system("pause")#define eps 1e-8#define PI acos(-1.0)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define maxn 100005#define lowbit(x) x&(-x)LL gcd(LL a,LL b) {return a?gcd(b%a,a):b;}int n;typedef struct hdu{    LL sum,cnt;}hdu;hdu C[maxn];void update(int pos,int val){    while(pos<maxn){        C[pos].cnt++;        C[pos].sum+=val;        pos+=lowbit(pos);    }}LL query1(int x){    LL ret=0;    while(x>0){        ret+=C[x].cnt;        x-=lowbit(x);    }    return ret;}LL query2(int x){    LL ret=0;    while(x>0){        ret+=C[x].sum;        x-=lowbit(x);    }    return ret;}int main(){    while(cin>>n){        clr(C,0);        LL sum=0,x,k1=0,k2=0;        for(int i=1;i<=n;i++){            cin>>x;            update(x,x);            k2=query2(maxn)-query2(x);            k1=i-query1(x);            if(k1!=0)            sum+=k1*x+k2;        }        printf("%I64d\n",sum);    }return 0;}