【HDOJ】1394 Minimum Inversion Number

逆序数的性质。
1. 暴力解

 #include <stdio.h>

 #define MAXNUM 5005

 int a[MAXNUM];

 int main() {
     int n;
     int i, j, sum, min;

     while (scanf("%d", &n) != EOF) {
         for (i=; i<n; ++i)
             scanf("%d", &a[i]);
         sum = ;
         for (i=; i<n; ++i) {
             for (j=i+; j<n; ++j)
                 if (a[j] < a[i])
                     ++sum;
         }
         min = sum;
         for (i=; i<n; ++i) {
             sum += n -  - a[i]*;
             if (sum < min)
                 min = sum;
             //printf("%d\n", sum);
         }
         printf("%d\n", min);
     }

     return ;
 }

2. 线段树。需要理解如何求逆序数。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;

 #define lson l, m, rt<<1
 #define rson m+1, r, rt<<1|1

 const int maxn = ;

 int nums[maxn<<];
 int a[maxn];

 void PushUP(int rt) {
     nums[rt] = nums[rt<<] + nums[rt<<|];
 }

 void build(int l, int r, int rt) {
     nums[rt] = ;
     if (l == r) return;
     int m = (l+r)>>;
     build(lson);
     build(rson);
 }

 void update(int p, int l, int r, int rt) {
     if (l == r) {
         nums[rt] = ;
         return ;
     }
     int m = (l+r)>>;
     if (p <= m)
         update(p, lson);
     else
         update(p, rson);
     PushUP(rt);
 }

 int query(int ll, int rr, int l, int r, int rt) {
     if (ll<=l && rr>=r) {
         return nums[rt];
     }
     int m = (l+r)>>;
     int ret = ;
     if (ll <= m)
         ret += query(ll, rr, lson);
     if (rr > m)
         ret += query(ll, rr, rson);

     return ret;
 }

 int main() {
     int n;
     int i, sum, min;

     while (scanf("%d", &n) != EOF) {
         build(, n-, );
         sum = ;
         for (i=; i<n; ++i) {
             scanf("%d", &a[i]);
             sum += query(a[i], n-, , n-, );
             update(a[i], , n-, );
             //printf("%d: sum=%d\n", i, sum);
         }
         min = sum;
         for (i=; i<n; ++i) {
             sum += n -  - a[i]*;
             if (sum < min)
                 min = sum;
         }
         printf("%d\n", min);
     }

     return ;
 }