1 /* 2 求逆序数的四种方法 3 */
1 /*
2 1. O(n^2) 暴力+递推 法:如果求出第一种情况的逆序列，其他的可以通过递推来搞出来，一开始是t[1],t[2],t[3]....t[N]
3 它的逆序列个数是N个，如果把t[1]放到t[N]后面，逆序列个数会减少t[1]个，相应会增加N-(t[1]+1)个
4 */
5 #include <cstdio>
6 #include <cstring>
7 #include <algorithm>
8 using namespace std;
9
10 const int MAX_N = 10000 + 10;
11 const int INF = 0x3f3f3f3f;
12 int a[MAX_N];
13 int num[MAX_N];
14
15 int main(void) //HDOJ 1394 Minimum Inversion Number
16 {
17 //freopen ("inC.txt", "r", stdin);
18 int n;
19
20 while (~scanf ("%d", &n))
21 {
22 memset (num, 0, sizeof (num));
23 for (int i=1; i<=n; ++i)
24 {
25 scanf ("%d", &a[i]);
26 //a[n+i] = a[i];
27 }
28 int t = 0; int sum = 0;
29 for (int i=1; i<=n; ++i) //先求解最初的数列逆序数
30 {
31 for (int j=i+1; j<=n; ++j)
32 {
33 if (a[i] > a[j])
34 {
35 sum++;
36 }
37 }
38 }
39 //printf ("%d\n", sum);
40 int ans = INF;
41 for (int i=1; i<=n; ++i) //更新sum，找最小
42 {
43 sum = sum - a[i] + (n - a[i] - 1); //the next line contains a permutation of the n integers from 0 to n-1.
44 //printf ("%d\n", res); //从0到n-1的整数 所以这里用a[i] 读题不仔细 。。。。
45 ans = min (sum, ans);
46 }
47 printf ("%d\n", ans);
48 }
49
50 return 0;
51 }
O(n^2) 暴力+递推
nlogn归并算法
1 /*
2 3. nlogn 线段树-单点更新：更新比a[i]大的个数
3 */
4 #include <cstdio>
5 #include <algorithm>
6 #define lson l, m, rt << 1
7 #define rson m+1, r, rt << 1 | 1
8
9 const int MAX_N = 5000 + 10;
10 const int INF = 0x3f3f3f3f;
11 int a[MAX_N];
12 int sum[MAX_N << 2];
13
14 void pushup(int rt)
15 {
16 sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
17 }
18
19 void build(int l, int r, int rt)
20 {
21 sum[rt] = 0;
22 if (l == r) return ;
23 int m = (l + r) >> 1;
24 build (lson);
25 build (rson);
26 }
27
28 void update(int p, int l, int r, int rt)
29 {
30 if (l == r)
31 {
32 sum[rt]++; //记录次数
33 return ;
34 }
35 int m = (l + r) >> 1;
36 if (p <= m)
37 {
38 update (p, lson);
39 }
40 else
41 update(p, rson);
42 pushup (rt);
43 }
44
45 int query(int ql, int qr, int l, int r, int rt)
46 {
47 if (ql <= l && r <= qr)
48 {
49 return sum[rt];
50 }
51 int m = (l + r) >> 1;
52 int ans = 0;
53 if (ql <= m) ans += query (ql, qr, lson);
54 if (qr > m) ans += query (ql, qr, rson);
55
56 return ans;
57 }
58
59 int main(void) //HDOJ 1394 Minimum Inversion Number
60 {
61 //freopen ("inC.txt", "r", stdin);
62 int n;
63 while (~scanf ("%d", &n))
64 {
65 //memset (num, 0, sizeof (num));
66 build (0, n-1, 1);
67 int sum = 0;
68 for (int i=1; i<=n; ++i)
69 {
70 scanf ("%d", &a[i]);
71 sum += query (a[i], n-1, 0, n-1, 1);
72 update (a[i], 0, n-1, 1);
73 }
74 int ans = sum;
75 for (int i=1; i<=n; ++i)
76 {
77 sum = sum - a[i] + (n - a[i] - 1);
78 ans = std::min (sum, ans);
79 }
80 printf ("%d\n", ans);
81 }
82
83 return 0;
84 }
nlogn 线段树-单点更新
1 /*
2 4. 树状数组
3 */
4 #include <stdio.h>
5 #include <string.h>
6 #include <algorithm>
7 using namespace std;
8 const int MAXN=5050;
9 int c[MAXN];
10 int a[MAXN];
11 int n;
12
13 int lowbit(int x)
14 {
15 return x&(-x);
16 }
17
18 void add(int i,int val)
19 {
20 while(i<=n)
21 {
22 c[i]+=val;
23 i+=lowbit(i);
24 }
25 }
26
27 int sum(int i)
28 {
29 int s=0;
30 while(i>0)
31 {
32 s+=c[i];
33 i-=lowbit(i);
34 }
35 return s;
36 }
37
38 int main() //HDOJ 1394 Minimum Inversion Number
39 {
40 //freopen ("inC.txt", "r", stdin);
41 while(scanf("%d",&n)!=EOF)
42 {
43 int ans=0;
44 memset(c,0,sizeof(c));
45 for(int i=1;i<=n;i++)
46 {
47 scanf("%d",&a[i]);
48 a[i]++;
49 ans+=sum(n)-sum(a[i]);
50 add(a[i],1);
51 }
52 int Min=ans;
53 for(int i=1;i<=n;i++)
54 {
55 ans+=n-a[i]-(a[i]-1);
56 if(ans<Min)Min=ans;
57 }
58 printf("%d\n",Min);
59 }
60
61 return 0;
62 }
树状数组
时间： 2024-10-12 03:24:17