原题地址
没仔细看题目。以为一个人只能选一次。。这样的话用线段树搞一下就可以，求出每个信号要抹去的最小代价，然后一个01背包就可以求出答案啦。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1e5+5;
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
LL a[N], Min[N<<2], sum, n, m, k, w[N];
LL dp[505];
void update(int rt, int l, int r, int L, int R, LL x) {
  if(l > R || L > r) return ;
  if(L <= l && R >= r) {
    Min[rt] = min(Min[rt], x);
    return ;
  }
  int mid = (l + r) / 2;
  update(lson, L, R, x);
  update(rson, L, R, x);
}
LL query(int rt, int l, int r, int pos) {
  if(l == r) return Min[rt];
  int mid = (l + r) / 2;
  if(pos <= mid) return min(Min[rt], query(lson, pos));
  else return min(Min[rt], query(rson, pos));
}
int main() {
  memset(Min, 0x3f, sizeof(Min));
  cin >> n >> k >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  int l, r;
  LL x;
  for (int i = 1; i <= m; i++) {
    cin >> l >> r >>x;
    update(1,1,n,l,r,x);
  }
  for (int i = 1; i <= n; i++) {
    w[i] = query(1,1,n,i);
  }
  for (int i = 1; i <= n; i++) {
    if(a[i] >= 0) continue;
    a[i] = -a[i];
    for (int j = k; j >= w[i]; j--) {
      dp[j] = max(dp[j], dp[j-w[i]]+a[i]);
    }
  }
  cout << sum + dp[k] << endl;
}