BZOJ 3333: 排队计划 树状数组+线段树

标签：bzoj   线段树   树状数组   逆序对   

题目大意：给出一个序列，求出这个序列的逆序对数量。定义一种操作，将一个数和他后面比他小的数字拿出来排序， 然后再放回去，之后输出逆序对数。

思路：思路题。手动模拟一下，会发现，逆序对变化的只是排序的那些点 。所以我们只要处理那些点就行了。先求一次逆序对，然后每次在拿出的数后面找到一个最小的数字，把它的权值改成INF，统计答案。

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 500010
#define INF 0x3f3f3f3f
using namespace std;
#define LEFT (pos << 1)
#define RIGHT (pos << 1|1)
#define min(a,b) ((a) < (b) ? (a):(b))
 
int cnt,asks;
pair<int,int *> xx[MAX];
int src[MAX],t;
int inv[MAX];
 
int fenwick[MAX];
long long now;
int tree[MAX << 2];
 
inline void Fix(int x)
{
    for(; x <= t; x += x&-x)
        ++fenwick[x];
}
 
inline int GetSum(int x)
{
    int re = 0;
    for(; x; x -= x&-x)
        re += fenwick[x];
    return re;
}
 
void BuildTree(int l,int r,int pos)
{
    if(l == r) {
        tree[pos] = l;
        return ;
    }
    int mid = (l + r) >> 1;
    BuildTree(l,mid,LEFT);
    BuildTree(mid + 1,r,RIGHT);
    tree[pos] = src[tree[LEFT]] < src[tree[RIGHT]] ? tree[LEFT]:tree[RIGHT];
}
 
int GetMin(int l,int r,int x,int y,int pos)
{
    if(l == x && y == r)
        return tree[pos];
    int mid = (l + r) >> 1;
    if(y <= mid) return GetMin(l,mid,x,y,LEFT);
    if(x > mid)      return GetMin(mid + 1,r,x,y,RIGHT);
    int left = GetMin(l,mid,x,mid,LEFT);
    int right = GetMin(mid + 1,r,mid + 1,y,RIGHT);
    return src[left] < src[right] ? left:right;
}
 
void Modify(int l,int r,int x,int pos,int c)
{
    if(l == r)  return ;
    int mid = (l + r) >> 1;
    if(x <= mid) Modify(l,mid,x,LEFT,c);
    else    Modify(mid + 1,r,x,RIGHT,c);
    tree[pos] = src[tree[LEFT]] < src[tree[RIGHT]] ? tree[LEFT]:tree[RIGHT];
}
 
int main()
{
    cin >> cnt >> asks;
    for(int i = 1; i <= cnt; ++i) {
        scanf("%d",&xx[i].first);
        xx[i].second = &src[i];
    }
    sort(xx + 1,xx + cnt + 1);
    for(int i = 1; i <= cnt; ++i) {
        if(i == 1 || xx[i].first != xx[i - 1].first)
            ++t;
        *xx[i].second = t;
    }
    for(int i = cnt; i; --i) {
        inv[i] = GetSum(src[i] - 1);
        now += inv[i];
        Fix(src[i]);
    }
    BuildTree(1,cnt,1);
    cout << now << endl;
    for(int temp,x,i = 1; i <= asks; ++i) {
        scanf("%d",&x);
        if(src[x] == INF) {
            printf("%lld\n",now);
            continue;
        }
        do {
            temp = GetMin(1,cnt,x,cnt,1);
            now -= inv[temp];
            Modify(1,cnt,temp,1,src[temp] = INF);
        }while(temp != x);
        printf("%lld\n",now);
    }
    return 0;
}

BZOJ 3333: 排队计划 树状数组+线段树

标签：bzoj   线段树   树状数组   逆序对   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/41775991