HDU - 1394 Minimum Inversion Number（树状数组 or 线段树）

题意：

给n个数字（0到n-1无重复），可以拿前面任意m个放到末尾。求拿多少个数字放到末尾后数列的逆序数最小。

解析：

逆序列个数是n个，如果把a[1]放到a
后面，逆序列个数会减少a[1]个，相应会增加n-(a[1]+1)个


比如说：

序列 3 6 9 0 8 5 7 4 2 1 把3移到后面，则它的逆序数会减少3个（0 2 1） 但同时会增加 n - a[i] - 1个。

AC代码

[code]#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 5*1e3 + 10;
int C
, n, a
;
int lowbit(int x) {
    return x & (-x);
}
int query(int x) {
    int ret = 0;
    while(x) {
        ret += C[x];
        x -= lowbit(x);
    }
    return ret;
}
void add(int x,int d) {
    while(x <= n) {
        C[x] += d;
        x += lowbit(x);
    }
}
int main() {
    while(scanf("%d", &n) != EOF) {
        memset(C, 0, sizeof(C));
        ll sum = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            add(a[i]+1, 1);
            sum += (a[i]+1 - query(a[i]+1));
        }
        ll ans = INF;
        for(int i = 1; i < n; i++) {
            sum = sum + (n-a[i]-1) - a[i];
            ans = min(ans, sum);
        }
        printf("%lld\n", ans);
    }
    return 0;
}

线段树方法求解：求解逆序数是没考虑清楚，将区间{a[i]+2, n}，错写成 {a[i]+1, n}，要吸取教训。

AC代码

[code]#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;
typedef long long ll;
const ll INF = 1e18;
const int N = 5e3 + 10;
int a
;
ll sum[N<<2];

void build(int o, int L, int R) {
    if(L == R) {
        sum[o] = a[L];
        return ;
    }
    int M = L + (R - L) / 2;
    build(o*2, L, M);
    build(o*2, M+1, R);
    sum[o] = sum[o*2] + sum[o*2+1];
}
int ql, qr;
int query(int o, int L, int R) {
    int M = L + (R - L) / 2, ans = 0;
    if(ql <= L && R <= qr) return sum[o];
    if(ql <= M) ans += query(o*2, L, M);
    if(M < qr) ans += query(o*2+1, M+1, R);
    return ans;
}
int p, v;
void modify(int o, int L, int R) {
    int M = L + (R - L) / 2;
    if(L == R) {
        sum[o] = v;
    }else {
        if(p <= M) modify(o*2, L, M);
        else modify(o*2+1, M+1, R);
        sum[o] = sum[o*2] + sum[o*2+1];
    }
}
int main() {
    int n;
    while(scanf("%d", &n) != EOF) {
        memset(sum, 0, sizeof(sum));
        build(1, 1, n);
        ll ans = 0;
        for(int i = 0; i < n; i++) {    
            scanf("%d", &a[i]);
            p = a[i]+1, v = 1;
            modify(1, 1, n);

            ql = a[i]+2, qr = n;
            ans += query(1, 1, n);
        }
        ll Min = ans;
        for(int i = 0; i < n; i++) {
            ans = ans + (n-a[i]-1) - a[i];
            Min = min(Min, ans);
        }
        printf("%lld\n", Min);
    }
    return 0;
}