思路：
离散化之后dp，dp[i][j]表示完成前i个任务并且处在第j个点所需要的最小代价。
实现：
1 #include <bits/stdc++.h>
2 using namespace std;
3 pair<int, int> a[1005];
4 int dp[1005][4005];
5 int main()
6 {
7 ios::sync_with_stdio(false);
8 int T, n; cin >> T;
9 while (T--)
10 {
11 cin >> n;
12 vector<int> v;
13 for (int i = 1; i <= n; i++)
14 {
15 cin >> a[i].first >> a[i].second;
16 v.push_back(a[i].first);
17 v.push_back(a[i].second);
18 if (a[i].first < a[i].second - 1)
19 {
20 v.push_back(a[i].first + 1);
21 v.push_back(a[i].second - 1);
22 }
23 }
24 sort(v.begin(), v.end());
25 v.erase(unique(v.begin(), v.end()), v.end());
26 int m = v.size();
27 for (int i = 1; i <= n; i++)
28 {
29 a[i].first = lower_bound(v.begin(), v.end(), a[i].first) - v.begin();
30 a[i].second = lower_bound(v.begin(), v.end(), a[i].second) - v.begin();
31 }
32 for (int i = 0; i < m; i++) dp[0][i] = 0;
33 for (int i = 1; i <= n; i++)
34 {
35 int l = a[i].first, r = a[i].second;
36 for (int j = l; j <= r; j++) dp[i][j] = dp[i - 1][j];
37 for (int j = 0; j < l; j++)
38 {
39 dp[i][j] = dp[i][l] + (abs(v[j] - v[l]) + 1 >> 1);
40 if (l < r) dp[i][j] = min(dp[i][j], dp[i][l + 1] + (abs(v[j] - v[l + 1]) + 1 >> 1));
41 }
42 for (int j = r + 1; j < m; j++)
43 {
44 dp[i][j] = dp[i][r] + (abs(v[j] - v[r] + 1) >> 1);
45 if (l < r) dp[i][j] = min(dp[i][j], dp[i][r - 1] + (abs(v[j] - v[r - 1]) + 1 >> 1));
46 }
47 }
48 cout << *min_element(dp[n], dp[n] + m) << endl;
49 }
50 return 0;
51 }