​​Culture Code​​

排完序之后dp。

dp[ i ] 表示 i 作为最里面一层的最小花费。

way[ i ] 表示得到最小花费的方案数。

然后用Map维护最小值就好了。 



//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);
 
using namespace std;
 
const int N = 5e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);
 
template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
int n;
int dp[N];
int way[N];
PII a[N];
 
priority_queue<pair<PII, PII>> q1;
map<int, int> Map;
 
int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d%d", &a[i].fi, &a[i].se);
    }
    sort(a + 1, a + 1 + n);
    reverse(a + 1, a + 1 + n);
    int maxIn = -inf;
    for(int i = 1; i <= n; i++) {
        int out = a[i].fi, in = a[i].se;
        if(maxIn < out) {
            dp[i] = 0;
            way[i] = 1;
        }
        else {
            while(SZ(q1) && q1.top().fi.fi >= out) {
                add(Map[q1.top().se.fi], q1.top().se.se);
                q1.pop();
            }
            assert(SZ(Map));
            auto t = *Map.begin();
            dp[i] = t.fi - out;
            way[i] = t.se;
        }
        chkmax(maxIn, in);
        q1.push(mk(mk(in, out), mk(dp[i] + in, way[i])));
    }
 
    int ans = inf;
 
    for(int i = 1; i <= n; i++) {
        dp[i] += a[i].se;
        chkmin(ans, dp[i]);
    }
 
    int ret = 0;
    for(int i = 1; i <= n; i++) {
        if(dp[i] == ans) {
            add(ret, way[i]);
        }
    }
 
    printf("%d\n", ret);
    return 0;
}
 
/*
*/