百度之星2017初赛B1006 小小粉丝度度熊

思路：

考虑到补签卡一定是连续放置才更优，所以直接根据起始位置枚举。预先处理区间之间的gap的前缀和，在枚举过程中二分即可。复杂度O(nlog(n))。

实现：

 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <algorithm>
 using namespace std;
 typedef long long ll;

 const ll INF = 0x3f3f3f3f;

 int main()
 {
     int n, m;
     while (scanf("%d %d", &n, &m) != EOF)
     {
         vector<pair<int, int>> v;
         int x, y;
         for (int i = ; i < n; i++)
         {
             scanf("%d %d", &x, &y);
             v.push_back(pair<int, int>(x, y));
         }
         sort(v.begin(), v.end());
         vector<pair<int, int>> res;
         res.push_back(v[]);
         for (int i = ; i < v.size(); i++)
         {
             if (v[i].first > res.back().second + ) res.push_back(v[i]);
             else res.back().second = max(res.back().second, v[i].second);
         }
         vector<ll> sum(res.size() - );
         for (int i = ; i < res.size() - ; i++)
             sum[i] = (ll)res[i + ].first -  - res[i].second;
         sum.push_back(INF);
         for (int i = ; i < sum.size(); i++)
             sum[i] += sum[i - ];
         ll maxn = ;
         for (int i = ; i < sum.size(); i++)
         {
             ll tmp = (i ==  ?  : sum[i - ]);
             int pos = upper_bound(sum.begin() + i, sum.end(), m + tmp) - sum.begin();
             ll tmp2 = (pos == i ? m : m - (sum[pos - ] - tmp));
             maxn = max(maxn, res[pos].second - res[i].first +  + tmp2);
         }
         printf("%lld\n", maxn);
     }
     return ;
 }