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 = 0; 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[0]);
        for (int i = 1; i < v.size(); i++)
        {
            if (v[i].first > res.back().second + 1) res.push_back(v[i]);
            else res.back().second = max(res.back().second, v[i].second);
        }
        vector<ll> sum(res.size() - 1);
        for (int i = 0; i < res.size() - 1; i++)
            sum[i] = (ll)res[i + 1].first - 1 - res[i].second;
        sum.push_back(INF);
        for (int i = 1; i < sum.size(); i++)
            sum[i] += sum[i - 1];
        ll maxn = 0;
        for (int i = 0; i < sum.size(); i++)
        {
            ll tmp = (i == 0 ? 0 : sum[i - 1]);
            int pos = upper_bound(sum.begin() + i, sum.end(), m + tmp) - sum.begin();
            ll tmp2 = (pos == i ? m : m - (sum[pos - 1] - tmp));
            maxn = max(maxn, res[pos].second - res[i].first + 1 + tmp2);
        }
        printf("%lld
", maxn);
    }
    return 0;
}