线段树 逆序对 Minimum Inversion Number HDU - 1394 Laptop

Minimum Inversion Number

HDU - 1394

求最小反转数，就是求最少的逆序对。

逆序对怎么求，就是先把所有的数都初始化为0，然后按照顺序放入数字，放入数字前查询从这个数往后面的数的位置是不是被占了，被占了说明有逆序对。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#define debug(n) printf("%d\n",n)
#define inf 0x3f3f3f3f;
using namespace std;
const int maxn =  + ;
int n, a[maxn];
struct node
{
    int l, r, num;
}tree[*maxn];

void push_up(int id)
{
    tree[id].num = tree[id << ].num + tree[id <<  | ].num;
}

void build(int id, int l, int r)
{
    tree[id].l = l;
    tree[id].r = r;
    if (l == r)
    {
        tree[id].num = ;
        return;
    }
    int mid = (l + r) >> ;
    build(id << , l, mid);
    build(id <<  | , mid + , r);
    push_up(id);
}

int query(int x, int y, int id)
{
    int l = tree[id].l;
    int r = tree[id].r;
    if (x <= l && y >= r)
    {
        return tree[id].num;
    }
    int mid = (l + r) >> ;
    int ans = ;
    if (x <= mid) ans += query(x, y, id << );
    if (y > mid) ans += query(x, y, id <<  | );
    push_up(id);
    return ans;
}

void update(int x, int y, int id)
{
    int l = tree[id].l;
    int r = tree[id].r;
    if (x <= l && y >= r)
    {
        tree[id].num = ;
        return;
    }
    int mid = (l + r) >> ;
    if (x <= mid) update(x, y, id << );
    if (y > mid) update(x, y, id <<  | );
    push_up(id);
}

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        int sum = ,ans=inf;
        build(, , n - );
        for (int i = ;i < n;i++)
        {

            scanf("%d", &a[i]);
            sum += query(a[i], n - , );
            update(a[i], a[i], );
            //debug(sum);
        }
        //debug(sum);
        for (int i = ;i < n;i++)
        {
            sum = sum + (n -  - a[i]) - a[i];
            ans = min(sum, ans);
        }
        printf("%d\n", ans);
    }
    return ;
}

Laptop

对其中一个进行排序，转化成逆序对。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#include <string>
#include <iostream>
#include <map>
#define inf 0x3f3f3f3f
#define inf64 0x3f3f3f3f3f3f3f3f
using namespace std;
const int maxn = 2e5 + ;
typedef long long ll;
map<ll, ll>mp;
struct node
{
    int l, r;
    int sum;
}tree[*maxn];
pair<ll, ll>a[maxn];
bool cmp(pair<ll,ll>a,pair<ll,ll>b)
{
    return a.first > b.first;
}
bool cmp1(pair<ll,ll>a,pair<ll,ll>b)
{
    return a.second < b.second;
}
void build(int id, int l, int r) {
    tree[id].l = l;
    tree[id].r = r;
    if (l == r) {
        tree[id].sum = ;
        return;
    }
    int mid = (l + r) >> ;
    build(id << , l, mid);
    build(id <<  | , mid + , r);
}

int query(int id,int x,int y)
{
    int l = tree[id].l;
    int r = tree[id].r;
    if(x<=l&&y>=r)
    {
    //    printf("id=%d sum=%d \n", id,tree[id].sum);
        return tree[id].sum;
    }
    int ans = ;
    int mid = (l + r) >> ;
    if (x <= mid) ans += query(id << , x, y);
    if (y > mid) ans += query(id <<  | , x, y);
//    printf("id=%d ans=%d l=%d r=%d x=%d y=%d \n", id, ans, l, r,x,y);
    return ans;
}

void push_up(int id)
{
    tree[id].sum = tree[id << ].sum + tree[id <<  | ].sum;
}

void update(int id,int x)
{
    int l = tree[id].l;
    int r = tree[id].r;
    if(l==r)
    {
        tree[id].sum = ;
        return;
    }
    int mid = (l + r) >> ;
    if (x <= mid) update(id << , x);
    else update(id <<  | , x);
    push_up(id);
}

int main()
{
    int n, tot = ;
    scanf("%d", &n);
    for (int i = ; i <= n; i++) scanf("%lld%lld", &a[i].first, &a[i].second);
    sort(a + , a +  + n, cmp1);
    for(int i=;i<=n;i++)
    {
        if (mp[a[i].second]) continue;
        mp[a[i].second] = ++tot;
    }
    sort(a + , a +  + n, cmp);
    build(, , tot);
    int ans = ;
    for(int i=;i<=n;i++)
    {
        if (query(, mp[a[i].second], tot)) ans++;
        update(, mp[a[i].second]);
    }
    printf("%d\n", ans);
    return ;
}
/*
 6
100 100
90 100
90 100
80 90
80 90
70 100
 */

还有一个是选拔赛的题目，学校oj崩了，下次再写。