树状数组

树状数组的区间求和单点更新只与操作数组c[] 相关 ; 与a[] 并无关系;

int lowbit(int x)
{
    return x & (-x);
}
void modify(int x,int add)  //一维(修改) ; 
{  
    while(x<=MAXN)  
    {      
        a[x]+=add;    
        x+=lowbit(x); 
    }
}
int get_sum(int x)      //求和 
{  
    int ret=0; 
    while(x!=0)  
    {       
        ret+=a[x];   
        x-=lowbit(x);   
    }  
    return ret;
}
void modify(int x,int y,int data)//二维(修改)
{
    for(int i=x;i<MAXN;i+=lowbit(i))
        for(int j=y;j<MAXN;j+=lowbit(j))
            a[i][j]+=data;
}
int get_sum(int x,int y)   //求和 
{
    int res=0;
    for(int i=x;i>0;i-=lowbit(i))
        for(int j=y;j>0;j-=lowbit(j))
            res+=a[i][j];
    return res;
}

int main()
{
    //1D; 
    {
        int a, b;
        printf("%d%d", &a, &b);
        modefy(a, b);
        sum=get_sum(b, a-1);    
    }
    //2D;
    {
        int x, x1, y, y1;
        scanf("%d%d%d%d", &x, &x1, &y, &y1);
        if(x>x1) swap(x, x1);
        if(y>y1) swap(y, y1);
        int rec=get_sum(x1, y1)-get_sum(x-1, y1)-get_sum(x1, y-1)+get_sum(x-1, y-1);
    }
    return 0;    
} 

hdoj2642--Stars

//注意数组脚标从1开始
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 1001
using namespace std;
int n, a[MAXN][MAXN]; bool b[MAXN][MAXN];
int lowbit(int x)
{
    return x &(-x);
}

void modify(int x, int y, int data)
{
    for(int i=x; i<=MAXN; i+= lowbit(i))
        for(int j=y; j<=MAXN; j+= lowbit(j))
            a[i][j] += data;
            
} 
int getSum(int x, int y)
{
    int res=0;
    for(int i=x; i>0; i -= lowbit(i))
        for(int j=y; j>0; j-= lowbit(j))
            res +=a[i][j];
    return res;
}
int main()
{
    scanf("%d", &n); 
    memset(b, 0, sizeof(b));
    while(n--)
    {
        int x, x1, y1, y; 
        char str[2]; scanf("%s", str);
        if(str[0]=='B')
        {
            scanf("%d%d", &x, &y);
            x++; y++;
            if(b[x][y]) continue;
            modify(x, y, 1);
            b[x][y]=1; 
        }
        else if(str[0]=='D')
        {
            scanf("%d%d", &x, &y);
            x++; y++;
            if(!b[x][y]) continue;
            modify(x, y, -1);
            b[x][y]=0;
        }
        else
        {
            scanf("%d%d%d%d", &x, &x1, &y, &y1);
            x++; y++; x1++; y1++;
            if(x>x1) swap(x, x1);
            if(y>y1) swap(y, y1);
            int rec=getSum(x1, y1)-getSum(x-1, y1)-getSum(x1, y-1)+getSum(x-1, y-1);
            printf("%d
", rec);
        }
        
    }
    return 0;
} 

hdoj1394求逆序数：

逆序对性质: 交换两个相邻数,逆序数+1或-1, 交换两个不相邻数a, b, 逆序数+=两者间大于a的个数-两者间小于a的个数

可得公式: ans= ans+n-a[i]-(a[i]-1);

#include <cstdio>
#include <cstring>
const int MAXN = 5050;
using namespace std;
int a[MAXN], c[MAXN];
int n;
int lowbit(int x)
{
    return x&(-x);
}

void add(int i, int val)
{
    while(i <=n)
    {
        c[i] += val;
        i +=lowbit(i);
    }
}

int sum(int i)
{
    int s=0;
    while(i>0)
    {
        s+=c[i];
        i-= lowbit(i);
    }
    return s;
}

int main()
{
    while(scanf("%d", &n) != EOF)
    {
        int ans=0;
        memset(c, 0, sizeof(c));
        for(int i=1; i<=n; i++)
        {
            scanf("%d", &a[i]);
            a[i]++;
            ans += sum(n)-sum(a[i]);
            add(a[i], 1);
        }
        int Min=ans;
        for(int i=1; i<=n; i++)
        {
            ans += n-a[i]-(a[i]-1);
            if(ans<Min) Min=ans;    
        } 
        printf("%d
", Min);
    }
    return 0;
}