HDU 1394 Minimum Inversion Number (线段树 求逆序对)

题目大意:

就是求逆序对之后O(1)递推每一种变化对应的逆序对数量的最小值


大致思路:

....刷


代码如下:

Result  :  Accepted     Memory  :  1644 KB     Time  :  93 ms

/*
 * Author: Gatevin
 * Created Time:  2015/8/14 22:01:45
 * File Name: Sakura_Chiyo.cpp
 */
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;

#define maxn 5050

struct Segment_Tree
{
    #define lson l, mid, rt << 1
    #define rson mid + 1, r, rt << 1 | 1
    int val[maxn << 2];
    void pushUp(int rt)
    {
        val[rt] = val[rt << 1] + val[rt << 1 | 1];
        return;
    }
    void build(int l, int r, int rt)
    {
        if(l == r)
        {
            val[rt] = 0;
            return;
        }
        int mid = (l + r) >> 1;
        build(lson);
        build(rson);
        pushUp(rt);
    }
    void update(int l, int r, int rt, int pos, int value)
    {
        if(l == r)
        {
            val[rt] += value;
            return;
        }
        int mid = (l + r) >> 1;
        if(mid >= pos) update(lson, pos, value);
        else update(rson, pos, value);
        pushUp(rt);
        return;
    }
    int query(int l, int r, int rt, int L, int R)
    {
        if(l >= L && r <= R)
            return val[rt];
        int mid = (l + r) >> 1;
        int ret = 0;
        if(mid >= L) ret += query(lson, L, R);
        if(mid + 1 <= R) ret += query(rson, L, R);
        return ret;
    }
};

Segment_Tree ST;
int a[maxn];

int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        ST.build(1, n + 1, 1);
        int sum = 0;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            a[i]++;
            sum += ST.query(1, n + 1, 1, a[i] + 1, n + 1);
            ST.update(1, n + 1, 1, a[i], 1);
        }
        int ans = sum;
        for(int i = 1; i <= n; i++)
        {
            sum += (n - a[i]);
            sum -= (a[i] - 1);
            ans = min(ans, sum);
        }
        printf("%d\n", ans);
    }
    return 0;
}



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