Codeforces 981 E - Addition on Segments

E - Addition on Segments

思路：

dp

dp[i]表示构成i的区间的右端点

先将询问按r排序

然后，对于每次询问，每次枚举 i 从 n-x 到 1，如果dp[i] >= l , 那么 dp[i+x] = max(dp[i+x], dp[x])

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 1e4 + 5;
struct node {
    int l, r, x;
    bool operator < (const node & t) const {
        return r < t.r;
    }
}Q[N];
int dp[N];
int main() {
    int n, q, l, r, x;
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= q; i++) {
         scanf("%d%d%d", &Q[i].l, &Q[i].r, &Q[i].x);
    }
    sort(Q+1, Q+1+q);
    for (int i = 1; i <= q; i++) {
        l = Q[i].l;
        r = Q[i].r;
        x = Q[i].x;
        for (int j = n-x; j >= 1; j--) {
            if(dp[j] >= l) dp[j+x] = max(dp[j+x], dp[j]);
        }
        dp[x] = r;
    }
    int ans = 0;
    for (int i = 1; i <= n; i++) ans += (dp[i]>0);
    printf("%d\n", ans);
    for (int i = 1; i <= n; i++) if(dp[i]) printf("%d ", i);
    return 0;
}