百度之星2019第一场1002 Game

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：

 #include <bits/stdc++.h>
 using namespace std;
 pair<int, int> a[];
 int dp[][];
 int main()
 {
     ios::sync_with_stdio(false);
     int T, n; cin >> T;
     while (T--)
     {
         cin >> n;
         vector<int> v;
         for (int i = ; i <= n; i++)
         {
             cin >> a[i].first >> a[i].second;
             v.push_back(a[i].first);
             v.push_back(a[i].second);
             if (a[i].first < a[i].second - )
             {
                 v.push_back(a[i].first + );
                 v.push_back(a[i].second - );
             }
         }
         sort(v.begin(), v.end());
         v.erase(unique(v.begin(), v.end()), v.end());
         int m = v.size();
         for (int i = ; i <= n; i++)
         {
             a[i].first = lower_bound(v.begin(), v.end(), a[i].first) - v.begin();
             a[i].second = lower_bound(v.begin(), v.end(), a[i].second) - v.begin();
         }
         for (int i = ; i < m; i++) dp[][i] = ;
         for (int i = ; i <= n; i++)
         {
             int l = a[i].first, r = a[i].second;
             for (int j = l; j <= r; j++) dp[i][j] = dp[i - ][j];
             for (int j = ; j < l; j++)
             {
                 dp[i][j] = dp[i][l] + (abs(v[j] - v[l]) +  >> );
                 if (l < r)
                     dp[i][j] = min(dp[i][j],
                         dp[i][l + ] + (abs(v[j] - v[l + ]) +  >> ));
             }
             for (int j = r + ; j < m; j++)
             {
                 dp[i][j] = dp[i][r] + (abs(v[j] - v[r] + ) >> );
                 if (l < r)
                     dp[i][j] = min(dp[i][j],
                         dp[i][r - ] + (abs(v[j] - v[r - ]) +  >> ));
             }
         }
         cout << *min_element(dp[n], dp[n] + m) << endl;
     }
     return ;
 }