Codeforces Round #301 (Div. 2) E. Infinite Inversions —— 逆序对离散化 + 树状数组

题目链接：http://codeforces.com/contest/540/problem/E

There is an infinite sequence consisting of all positive integers in the increasing order: p = {1, 2, 3, ...}. We performed n swap operations with this sequence. A swap(a, b) is an operation of swapping the elements of the sequence on positions a and b. Your task is to find the number of inversions in the resulting sequence, i.e. the number of such index pairs (i, j), that i < j and pi > pj.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of swap operations applied to the sequence.

Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 109, ai ≠ bi) — the arguments of the swap operation.

Output

Print a single integer — the number of inversions in the resulting sequence.

Examples

input

Copy

2
4 2
1 4

output

Copy

input

Copy

output

Copy

Note

In the first sample the sequence is being modified as follows: $\text{[math]}$ . It has 4 inversions formed by index pairs (1, 4), (2, 3), (2, 4) and (3, 4).

方法一：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <cstdlib>
 6 #include <string>
 7 #include <vector>
 8 #include <map>
 9 #include <set>
10 #include <queue>
11 #include <sstream>
12 #include <algorithm>
13 using namespace std;
14 typedef long long LL;
15 const double eps = 1e-8;
16 const int INF = 2e9;
17 const LL LNF = 9e18;
18 const int MOD = 1e9+7;
19 const int MAXN = 2e5+10;
20 
21 int c[MAXN];
22 int lowbit(int x)
23 {
24     return x&(-x);
25 }
26 
27 void add(int x, int d)
28 {
29     for(; x<MAXN; x += lowbit(x))
30         c[x] += d;
31 }
32 
33 int sum(int x)
34 {
35     int ret = 0;
36     for(; x>0; x -= lowbit(x))
37         ret += c[x];
38     return ret;
39 }
40 
41 int query(int l, int r)
42 {
43     if(l>r) return 0;
44     return sum(r)-sum(l-1);
45 }
46 
47 int x[MAXN], y[MAXN], a[MAXN], pre[MAXN], now[MAXN];
48 map<int,int>M;
49 int main()
50 {
51     int n;
52     while(scanf("%d",&n)!=EOF)
53     {
54         M.clear();
55         for(int i = 1; i<=n; i++)
56         {
57             scanf("%d%d",&x[i],&y[i]);
58             a[i] = x[i]; a[n+i] = y[i];
59         }
60         sort(a+1,a+1+2*n);
61         int m = unique(a+1,a+1+2*n)-(a+1);
62         memcpy(pre+1, a+1, m*sizeof(a[1]));
63         memcpy(now+1, a+1, m*sizeof(a[1]));
64         for(int i = 1; i<=m; i++)
65             M[now[i]] = i;
66 
67         for(int i = 1; i<=n; i++)
68         {
69             swap(now[M[x[i]]], now[M[y[i]]]);
70             swap(M[x[i]], M[y[i]]);
71         }
72 
73         LL ans = 0;
74         memset(c, 0, sizeof(c));
75         for(int i = 1; i<=m; i++)
76         {
77             int pre_pos = upper_bound(pre+1,pre+1+m, now[i])-(pre+1);
78             int now_pos = i;
79             ans += query(pre_pos+1,m);
80             add(pre_pos, 1);
81 
82             if(pre[i]>now[i]) ans += (pre[i]-now[i])-(now_pos-pre_pos);
83             if(pre[i]<now[i]) ans += (now[i]-pre[i])-(pre_pos-now_pos);
84         }
85         printf("%lld\n",ans);
86     }
87 }

View Code

方法二：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <cstdlib>
 6 #include <string>
 7 #include <vector>
 8 #include <map>
 9 #include <set>
10 #include <queue>
11 #include <sstream>
12 #include <algorithm>
13 using namespace std;
14 typedef long long LL;
15 const double eps = 1e-8;
16 const int INF = 2e9;
17 const LL LNF = 9e18;
18 const int MOD = 1e9+7;
19 const int MAXN = 4e5+10;
20 
21 struct node
22 {
23     int val, cnt;
24 }b[MAXN];
25 
26 int c[MAXN];
27 int lowbit(int x)
28 {
29     return x&(-x);
30 }
31 
32 void add(int x, int d)
33 {
34     for(; x<MAXN; x += lowbit(x))
35         c[x] += d;
36 }
37 
38 int sum(int x)
39 {
40     int ret = 0;
41     for(; x>0; x -= lowbit(x))
42         ret += c[x];
43     return ret;
44 }
45 
46 int query(int l, int r)
47 {
48     if(l>r) return 0;
49     return sum(r)-sum(l-1);
50 }
51 
52 int x[MAXN], y[MAXN], a[MAXN];
53 map<int,int>M, MM;
54 int main()
55 {
56     int n;
57     while(scanf("%d",&n)!=EOF)
58     {
59         M.clear();
60         MM.clear();
61         for(int i = 1; i<=n; i++)
62         {
63             scanf("%d%d",&x[i],&y[i]);
64             a[i] = x[i]; a[n+i] = y[i];
65         }
66         sort(a+1,a+1+2*n);
67         int m = unique(a+1,a+1+2*n)-(a+1);
68         int cnt = 0, st = 1;
69         for(int i = 1; i<=m; i++)
70         {
71             if(st!=a[i])
72             {
73                 b[++cnt].val = st;
74                 b[cnt].cnt = a[i]-st;
75                 M[b[cnt].val] = MM[b[cnt].val] = cnt;
76             }
77             b[++cnt].val = a[i];
78             b[cnt].cnt = 1;
79             M[b[cnt].val] = MM[b[cnt].val] = cnt;
80             st = a[i]+1;
81         }
82         m = cnt;
83 
84         for(int i = 1; i<=n; i++)
85         {
86             swap(b[MM[x[i]]], b[MM[y[i]]]);
87             swap(MM[x[i]], MM[y[i]]);
88         }
89 
90         memset(c, 0,sizeof(c));
91         LL ans = 0;
92         for(int i = 1; i<=m; i++)
93         {
94             ans += 1LL*b[i].cnt*query(M[b[i].val]+1, m);
95             add(M[b[i].val], b[i].cnt);
96         }
97         printf("%lld\n",ans);
98     }
99 }

View Code

Codeforces Round #301 (Div. 2) E. Infinite Inversions —— 逆序对 离散化 + 树状数组

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Codeforces Round #301 (Div. 2) E. Infinite Inversions —— 逆序对离散化 + 树状数组