hdu 1394 Minimum Inversion Number（线段树）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1394

题意：给定一个全排列，这道题可以看作求该循环序列的最小逆序数。

思路：

数据范围只有5000，直接暴力就可以过，187ms。

但是可以用线段树来优化，46ms可以过。

这道题需要用到一个结论，将一个数移动到序列的最后，逆序数增加(-x+n-1-x)。

具体解释一下，去掉一个数，假设该数的逆序数为x，则该序列的逆序数减少x，意味着在该数之后有x个数比他小，

然后把这个数添加到最后面，在该数后面的数有n-1个数中有n-1-x个数比他大，所以序列的逆序数会相应的增加n-1-x.

暴力：

#include <limits.h>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <time.h>#include <algorithm>#include <iostream>#include <iterator>#include <sstream>#include <queue>#include <stack>#include <string>#include <vector>#include <list>#include <set>//#define ONLINE_JUDGE#define eps 1e-6#define INF 0x7fffffff                                          //INT_MAX#define inf 0x3f3f3f3f                                          //int??????????????????#define FOR(i,a) for((i)=0;i<(a);(i)++)                          //[i,a);#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define MEM3(a) memset(a,0x3f,sizeof(a))#define MEMS(a) memset(a,'\0',sizeof(a))#define LL __int64const double PI = acos(-1.0);template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }template<class T> inline T Min(T a, T b) { return a < b ? a : b; }template<class T> inline T Max(T a, T b) { return a > b ? a : b; }using namespace std;template<class T>T Mint(T a, T b, T c) {    if (a>b) {        if (c>b)            return b;        return c;    }    if (c>a)        return a;    return c;}template<class T>T Maxt(T a, T b, T c) {    if (a>b) {        if (c>a)            return c;        return a;    }    else if (c > b)        return c;    return b;}const int Maxn=500010;const int maxn=5005;int T,n,m;int a[maxn];int main(){#ifndef ONLINE_JUDGE    freopen("test.in","r",stdin);    freopen("test.out","w",stdout);#endif    while(~sf(n)){        for1(i,0,n) sf(a[i]);        int cnt=0,ans=inf;        for1(i,0,n){            for1(j,i+1,n){                if(a[i]>a[j])                    cnt++;            }        }        if(ans>cnt) ans=cnt;        for1(i,0,n){            cnt=cnt-a[i]+n-1-a[i];            if(ans>cnt)                ans=cnt;        }        pf(ans);    }    return 0;}

线段树：

#include <limits.h>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <time.h>#include <algorithm>#include <iostream>#include <iterator>#include <sstream>#include <queue>#include <stack>#include <string>#include <vector>#include <list>#include <set>//#define ONLINE_JUDGE#define eps 1e-6#define INF 0x7fffffff                                          //INT_MAX#define inf 0x3f3f3f3f                                          //int??????????????????#define FOR(i,a) for((i)=0;i<(a);(i)++)                          //[i,a);#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define MEM3(a) memset(a,0x3f,sizeof(a))#define MEMS(a) memset(a,'\0',sizeof(a))#define LL __int64const double PI = acos(-1.0);template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }template<class T> inline T Min(T a, T b) { return a < b ? a : b; }template<class T> inline T Max(T a, T b) { return a > b ? a : b; }using namespace std;template<class T>T Mint(T a, T b, T c) {    if (a>b) {        if (c>b)            return b;        return c;    }    if (c>a)        return a;    return c;}template<class T>T Maxt(T a, T b, T c) {    if (a>b) {        if (c>a)            return c;        return a;    }    else if (c > b)        return c;    return b;}const int Maxn=500010;const int maxn=5005;int T,n,m;int a[maxn];struct segTree{    int l,r;    int v;}node[3*maxn];void pushup(int id){    node[id].v=node[id<<1].v+node[id<<1|1].v;}void build(int id,int l,int r){    node[id].l=l;    node[id].r=r;    node[id].v=0;    if(l==r)        return ;    int mid=(l+r)>>1;    build(id<<1,l,mid);    build(id<<1|1,mid+1,r);}void Update(int id,int pos){    int L=node[id].l,R=node[id].r;    if(L==R){        node[id].v++;        return ;    }    int mid=(L+R)>>1;    if(pos<=mid) Update(id<<1,pos);    else Update(id<<1|1,pos);    pushup(id);}int Query(int id,int p,int q){    int L=node[id].l,R=node[id].r;    if(p<=L&&R<=q) return node[id].v;    int mid=(L+R)>>1;    if(q<=mid) return Query(id<<1,p,q);    if(p>mid) return Query(id<<1|1,p,q);    return Query(id<<1,p,mid)+Query(id<<1|1,mid+1,q);}int main(){#ifndef ONLINE_JUDGE    freopen("test.in","r",stdin);    freopen("test.out","w",stdout);#endif    while(~sf(n)){        build(1,0,n-1);        int ans=inf,cnt=0;        for1(i,0,n) {            sf(a[i]);            int num=Query(1,a[i]+1,n-1);            cnt+=num;            Update(1,a[i]);        }        if(ans>cnt) ans=cnt;        for1(i,0,n){            cnt=cnt-a[i]+n-1-a[i];            if(ans>cnt) ans=cnt;        }        pf(ans);    }    return 0;}