求逆序对

POJ - 2299

题意就是求逆序对，当然还有别的方法，如果之后写了会补充

树状数组：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
#define ll long long
const int maxn = 5e5 + 50;
int d[maxn],id[maxn];
ll sum[maxn];
int lowbit(int x){return x&-x;}
int num;
void add(int id)
{
    for(int i = id; i <= num ; i += lowbit(i))  sum[i] += 1;
}
ll query(int id)
{
    ll cnt = 0;
    for(int i = id; i ; i -= lowbit(i)) cnt += sum[i];
    return cnt;
}
int n;
int main()
{
    while(~scanf("%d",&n) && n)
    {
        memset(sum,0,sizeof(sum));

        for(int i = 0; i < n; i++) scanf("%d",&d[i]), id[i+1] = d[i];
        sort(id + 1,id + n + 1);
        num = unique(id+1,id+1+n) - id;
        ll ans = 0;
        for(int i = 0; i < n; i++)
        {
            int th = lower_bound(id + 1,id + num,d[i])- id;
            ll temp = query(num) - query(th);
            ans += temp;
            add(th);
        }
        printf("%lld\n",ans);
    }
	return 0;
}

线段树：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
#define ll long long
const int maxn = 5e5 + 50;
ll sum[maxn<<2];
int d[maxn], id[maxn];
int n,num;
void add(int id,int l,int r,int u)
{
    if(l==r && l == u)
    {
        sum[id] ++;
        return;
    }
    int mid = l+r>>1;
    if(u <= mid) add(id<<1,l,mid,u);
    else add(id << 1 |1 ,mid+1,r,u);

    sum[id] = sum[id<<1] + sum[id<<1|1];
}

ll query(int id,int l,int r,int ql,int qr)
{
    if(l > qr || r < ql) return 0;
    if(l >= ql && r <= qr) return sum[id];

    int mid = l+r>>1;
    ll ans = 0;
    if(ql <= mid) ans += query(id<<1,l,mid,ql,qr);
    if(qr > mid) ans += query(id<<1|1,mid+1,r,ql,qr);
    return ans;
}


int main()
{
    while(~scanf("%d",&n) && n)
    {
        memset(sum,0,sizeof(sum));
        for(int i = 0; i < n; i++)
            scanf("%d",&d[i]), id[i+1] = d[i];
        sort(id + 1 , id + n + 1);
        num = unique(id + 1 , id + n + 1) - id;
        ll ans = 0;
        for(int i = 0; i < n; i++)
        {
            int th = lower_bound(id+1,id+num,d[i]) - id;
            ll temp = query(1,1,num,th+1,num);
            ans += temp;
            add(1,1,num,th);
        }
        printf("%lld\n",ans);
    }
	return 0;
}