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\(DP\)神题。以后要多学习一个，练一练智商。

关键点在于把“有\(a_i\)个人分数比我高，\(b_i\)个人分数比我低”这句话转换成“排名为\(a_i+1\)，且有\(n-a_i-b_i\)个人和我分数相同“。解决了这一点，问题就解决了一大半，接下来就变成了最大不相交区间集合选择问题。本来我是用最长路写的，不知道为什么出锅了，所以就改用\(DP\)+二分了。

#include <bits/stdc++.h>
using namespace std;

const int N = 100010;

struct Segment {
    int l, r, sz;
    
    Segment (int _l = 0, int _r = 0, int _sz = 0) {
        l = _l, r = _r, sz = _sz;
    }

    bool operator < (Segment rhs) const {return l == rhs.l ? r < rhs.r : l < rhs.l;}
    bool operator == (Segment rhs) const {return l == rhs.l && r == rhs.r;}
} arr[N];

map <Segment, int> mp;

int n, a[N], b[N], tot, dp[N], dis[N], vis[N];

bool cmp (Segment lhs, Segment rhs) {return lhs.r == rhs.r ? lhs.l < rhs.l : lhs.r < rhs.r;}

int main () {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i] >> b[i];
        if (a[i] + b[i] <= n - 1) {
            arr[++tot] = Segment (a[i] + 1, n - b[i], n - a[i] - b[i]);
            mp[arr[tot]]++;
        }
    }
    sort (arr + 1, arr + 1 + tot);
    tot = unique (arr + 1, arr + 1 + tot) - arr - 1;
    for (int i = 1; i <= tot; ++i) {
        mp[arr[i]] = min (mp[arr[i]], arr[i].sz);
        // printf ("Segment %d = [%d, %d]\n", i, arr[i].l, arr[i].r);
    }
    sort (arr + 1, arr + 1 + tot, cmp);
    for (int i = 1; i <= tot; ++i) {
        dp[i] = max (dp[i], dp[i - 1]);
        int l = 1, r = i;
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (arr[mid].r < arr[i].l) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        dp[i] = max (dp[i], dp[r] + mp[arr[i]]);
    }
    cout << n - dp[tot] << endl;
}