POJ 3616 Milking Time——区间DP

一共有M个区间，我们可以对这M个区间进行dp，定义dp[i]为遍历前i - 1个区间得到的最大产奶量(要保证【第i个区间的开始时间】小于【从前i-1个区间选出的一个区间的结束时间】),注意要先对区间的时间进行排序，还要注意dp[M]并不一定是结果，因为可能根本达不到dp[M],最后要对所有dp值遍历一便找最大值

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn = 1000000 + 10;

int dp[maxn];
struct Time {
int left, right, val;
bool operator < (const Time &another) const {
if (left != another.left) return left < another.left;
else return right < another.right;
}
}time[maxn];

int main()
{
int N, M, R;
scanf("%d %d %d", &N, &M, &R);
for (int i = 1; i <= M; i++) {
scanf("%d %d %d", &time[i].left, &time[i].right, &time[i].val);
time[i].right += R;
}
sort(time + 1, time + 1 + M);
for (int i = 1; i <= M; i++) {
dp[i] = time[i].val;
for (int j = 1; j < i; j++) {
if (time[j].right <= time[i].left) {
dp[i] = max(dp[i], dp[j] + time[i].val);
}
}
}
int ans = -1;
for (int i = 1; i <= M; i++) ans = max(dp[i], ans);
cout << ans << endl;
return 0;
}