E. Infinite Inversions
time limit per test
2 seconds
memory limit per test
256 megabytes
input standard input
output standard output
There is an infinite sequence consisting of all positive integers in the increasing order: p = {1, 2, 3, ...}. We performed n swapoperations 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.
Sample test(s)
input
24 21 4
output
4
input
31 63 42 5
output
15
题意：一个无限 长的数组（1， 2， 3， 4， .....）， n次操作， 每次交换两个位置上的值.
输出最终 有多少逆序对数。
由于这题是 数字可能会很多，，我们只能 离散化之后来求 逆序数了，， 先把所有操作读进来，，离散化 被操作数。 而那些被操作数之间的数字可以缩点来处理（就是把所有的数字看成一个数字来处理）。然后就可以求出结果了。。
1 #include <set>
2 #include <map>
3 #include <cmath>
4 #include <ctime>
5 #include <queue>
6 #include <stack>
7 #include <cstdio>
8 #include <string>
9 #include <vector>
10 #include <cstdlib>
11 #include <cstring>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 typedef unsigned long long ull;
16 typedef long long ll;
17 const int inf = 0x3f3f3f3f;
18 const double eps = 1e-8;
19 const int MAXN = 4e5+10;
20 int a[MAXN], tot, n;
21 int A[MAXN], B[MAXN];
22 int lowbit (int x)
23 {
24 return x & -x;
25 }
26 long long arr[MAXN], M;
27 void modify (int x, int d)
28 {
29 while (x < M)
30 {
31 arr[x] += d;
32 x += lowbit (x);
33 }
34 }
35 int sum(int x)
36 {
37 int ans = 0;
38 while (x)
39 {
40 ans += arr[x];
41 x -= lowbit (x);
42 }
43 return ans;
44 }
45 int p[MAXN],kk[MAXN];
46 int main()
47 {
48 #ifndef ONLINE_JUDGE
49 freopen("in.txt","r",stdin);
50 #endif
51 while (cin >> n)
52 {
53 int x, y;
54 tot = 0;
55 memset (arr, 0, sizeof (arr));
56 memset(kk, 0, sizeof (kk));
57 long long minv = inf;
58 long long maxv = 0;
59 map<int, int>pp;
60 for (int i = 0; i < n; i++)
61 {
62 scanf ("%d%d", &x, &y);
63 minv = min(minv, (long long)min(x, y));
64 maxv = max(maxv, (long long)max(x, y));
65 a[tot++] = x;
66 a[tot++] = y;
67 A[i] = x;
68 B[i] = y;
69 }
70 sort (a, a+tot);
71
72 tot = unique(a, a+tot) - a;
73 int ok = 0;
74 int tmp = tot;
75 long long j = minv;
76 int tt;
77 vector<int>vec;
78 for (int i = 0; i < tot; )
79 {
80 if (a[i] == j)
81 {
82 i++;
83 j++;
84 ok = 0;
85 }
86 else
87 {
88 if (ok == 0) // 缩点
89 {
90 ok = j;
91 a[tmp++] = j;
92 tt = j;
93 vec.push_back(tt);
94 }
95 pp[ok] += a[i]-j;
96 j = a[i];
97 }
98 }
99 tot = tmp;
100 sort (a, a+tot);
101 for (int i = 0; i < vec.size(); i++)
102 {
103 int qq = vec[i];
104 if (pp.count(qq) >= 1)
105 {
106 int ix = lower_bound(a, a+tot, qq)-a+1;
107 kk[ix] = pp[qq];
108 }
109 }
110 for (int i = 0; i < n; i++)
111 {
112 A[i] = lower_bound(a,a+tot, A[i]) - a + 1; // 离散化
113 B[i] = lower_bound(a,a+tot, B[i]) - a + 1;
114 }
115 maxv = lower_bound(a, a+tot, maxv) - a + 1;
116 M = maxv+10;
117 for (int i = 1; i <= maxv; i++)
118 {
119 p[i] = i;
120 }
121 for (int i = 0; i < n; i++)
122 {
123 swap(p[A[i]], p[B[i]]);
124 }
125 long long ans = 0;
126 long long cnt = 0;
127 for (int i = 1; i <= maxv; i++)
128 {
129 ans += (long long)(i-1+cnt - sum(p[i]))*max(1,kk[i]);
130 modify(p[i], max(1,kk[i]));
131 cnt += max(1,kk[i]) - 1;
132 }
133 printf("%I64d\n", ans);
134 }
135 return 0;
136 }