这个dp题很有学问,我也是照着标称写的
还需要学习
补: if(order[i] < order[i-1]) pre[j] += now[j]; 这句的解释

首先order表示的是每个数字排序之后的数组
order[0] 就是最小的那个数字是原来数组哪一个

但是会有等于的情况 所以这里定义的是如果
i < j时
* order[i] > order[j] 意味 A{order[i]] >= A[order[j]]
* order[j] < order[j] 意味 A[order[i]] < A[order[j]]

#include<vector>

#include<iostream>

#include<cstdio>

#include<algorithm>

using namespace std;

const int MAXN = 1e5+5;

typedef long long ll;

int main(){

    int n;

    while(~scanf("%d",&n)) {

        vector<int> low(n), high(n);

        int m = 0;

        for(int i = 0; i < n; ++i) {

            scanf("%d %d",&low[i],&high[i]);

            m = max(m, high[i]+1);

        }

        vector<ll> result(n+1);

        vector<int> order(n);

        for(int i = 0; i < n; ++i) order[i] = i;

        do{

            vector<ll> now(m);

            for(int i = low[order[0]]; i <= high[order[0]]; ++i) now[i] ++;

            for(int i = 1; i < n; ++i) {

                vector<ll> pre(m);

                ll sum = 0;

                int lo = low[order[i]]; int hi = high[order[i]];

                for(int j = 0; j < m; ++j) {

                    if(lo <= j && j <= hi) {

                        pre[j] = sum;

                        if(order[i] < order[i-1]) pre[j] += now[j];

                    }

                    sum += now[j];

                }

                now.swap(pre);

            }

            vector<int> dp(n,1);

            for(int i = 0; i < n; ++i)

                for(int j = 0; j < i; ++j)

                    if(order[j] < order[i]) {

                        dp[i] = max(dp[i], dp[j]+1);

                    }

            int maxx = *max_element(dp.begin(), dp.end());

            for(int i = 0; i < m; ++i)

                result[maxx] += now[i];

        }while(next_permutation(order.begin(), order.end()));

        for(int i = 1; i <= n; ++i) {

            if(i!=1) printf(" ");

            printf("%lld",result[i]);

        } printf("\n");

    }

    return 0;

}