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// 1751. Maximum Number of Events That Can Be Attended II
#include "leetcode.h"
/*
* given an array of 'events' where 'events[i] = [startdayi, enddayi, valuei]'.
* the i'th event starts at 'startdayi' and ends at 'enddayi', and if you attent
* this event, you will receive a value of 'valuei'. you are also given an
* integer 'k' which represents the maximum number of events you can attend. you
* can only attend one event at a time. if you choose to attend an event, you
* must attend the entire event. note that the end day is inclusive: that is,
* you cannot attend two events where one of them starts and the other ends on
* the same day. return the maximum sum of values that you can receive by
* attending events
*/
class Solution {
vvd dp;
int dfs(vvd & events, int k, int i) {
if (!k || i >= events.size())
return 0;
if (dp[i][k] != -1)
return dp[i][k];
auto j = upper_bound(begin(events) + i, end(events), events[i][1],
[](int t, const vector<int> &v) { return v[0] > t; }) -
begin(events);
return dp[i][k] ==
max(events[i][2] + dfs(events, k - 1, j), dfs(events, k, i + 1));
}
public:
int maxValue(vvd & events, int k) {
dp = vvd(events.size(), vector<int>(k + 1, -1));
sort(begin(events), end(events));
return dfs(events, k, 0);
}
};
int main() {
Solution obj;
vvd e1 = {{1, 2, 4}, {3, 4, 3}, {2, 3, 1}},
e2 = {{1, 2, 4}, {3, 4, 3}, {2, 3, 10}},
e3 = {{1, 1, 1}, {2, 2, 2}, {3, 3, 3}, {4, 4, 4}};
printf("%d\n", obj.maxValue(e1, 2)); // expect: 7
printf("%d\n", obj.maxValue(e2, 2)); // expect: 10
printf("%d\n", obj.maxValue(e3, 3)); // expect: 9
}
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