很久就过掉了，补个题解。
先思考：设(dp[i])为时刻(i)获得的最大空闲时间，那么：
[dp[i] =
egin{cases}
dp[i - 1] + 1 & exttt{don't have work}\
max {dp[i + T[s]} & exttt{have work at } color{red}{ exttt{s}}
end{cases}
]
会发现，(dp[i + T[s]])需要先计算，所以要从(n)到(1)计算，递推式也要改一改：
[dp[i] =
egin{cases}
dp[i + 1] + 1 & exttt{don't have work}\
max {dp[i + T[s]} & exttt{have work at } color{red}{ exttt{s}}
end{cases}
]
时间复杂度显然是(O(n))的
#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 5;
int dp[N];
int n, k;
vector<int> T[N];
int main(void)
{
scanf("%d%d", &n, &k);
for (int i = 1; i <= k; i++)
{
int s, t;
scanf("%d%d", &s, &t);
T[s].push_back(t);
}
for (int i = n; i >= 1; i--)
{
if (T[i].size() == 0)
dp[i] = dp[i + 1] + 1;
else
{
for (int j = 0; j < T[i].size(); j++)
dp[i] = max(dp[i], dp[i + T[i][j]]);
}
}
printf("%d
", dp[1]);
return 0;
}