hdu1394线段树点修改，区间求和

Posted 2020-09-03 starry

Problem Description

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

 

 

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

 

 

Output

For each case, output the minimum inversion number on a single line.

 

 

Sample Input

10 1 3 6 9 0 8 5 7 4 2

 

 

Sample Output

16

 

一开始看这题时，弄好好久才把题目意思弄懂。

 

 

题目意思是求最小逆序数，而逆序数就是前面的数大于后面的数，而一串数字又可以移动，现在要求逆序数最小的是有多少。

 

我们可以先把每个数的当前位置有多少个位置求出来，如a1,a2,a3,a4......an分别对应有b1,b2,b3,b4......bn个逆序数。求出一开始的逆序数sn，但把第一个数放在最后时：

a2,a3,a4......an,a1  这时的逆序数就sn - a1 + n - a1 - 1 保留最小值即可，然后遍历整个数组就可以求出最小逆序数了。

下面贴下我写的 线段树+暴力 代码

 

线段树代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #define lson l, m, rt<<1
 4 #define rson m+1, r, rt<<1|1
 5 using namespace std;
 6 const int maxn = 5e3+10;
 7 
 8 int min(int a, int b){
 9     return a>b?b:a;
10 }
11 int sum[maxn<<2], a[maxn];
12 void PushUP(int rt){
13     sum[rt] = sum[rt<<1] + sum[rt<<1|1];
14 }
15 void build(int l, int r, int rt){
16     sum[rt] = 0;
17     if(l == r)return;
18     int m = (l + r) >> 1;
19     build(lson);
20     build(rson);
21 }
22 int query(int L, int R, int l, int r, int rt){
23     if(L <= l && r <= R){
24         return sum[rt];
25     }
26     int m = (l + r) >> 1;
27     int ans = 0;
28     if(m >= L) ans +=query(L, R, lson);
29     if(m < R) ans += query(L, R, rson);
30     return ans;
31 }
32 void update(int p, int l, int r, int rt){
33     if(l == r){
34         sum[rt]++;
35         return;
36     }
37     int m = (l + r) >> 1;
38     if(p <= m) update(p,lson);
39     else update(p,rson);
40     PushUP(rt);
41 }
42 int main(){
43     int n;
44     while(cin>>n){
45         int sn = 0;
46         build(0, n - 1, 1);
47         for(int i = 0; i < n; i++){
48             scanf("%d",&a[i]);
49             sn += query(a[i], n - 1, 0, n -1, 1);
50             update(a[i], 0, n - 1, 1);
51         }
52         int ans = sn;
53         for(int i = 0; i < n; i++){
54             sn = sn - a[i] + n - a[i] - 1;
55             if(sn < ans){
56                 ans = sn;
57             }
58         }
59         cout << ans << endl;
60     }
61     return 0;
62 }

 

 

 

暴力代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 int a[5005],b[5005];
 6 int main(){
 7     int n;
 8     while(cin>>n){
 9         memset(a, 0, sizeof(a));
10         memset(b, 0, sizeof(b));
11         for(int i = 0; i < n; i++){
12             scanf("%d",&a[i]);
13         }
14         int sum = 0;
15         for(int i = 0; i < n; i++){
16             for(int j = i+1; j < n; j++){
17                 if(a[i] > a[j]){
18                     b[i]++;
19                 }
20             }
21             sum += b[i];
22         }
23         int minx = sum;
24         for(int i = 0; i < n; i++){
25             sum = sum - a[i] + (n - a[i] - 1);
26             if(sum < minx){
27                 minx = sum;
28             }
29         }
30         cout << minx << endl;
31     }
32     return 0;
33 }