P1280 尼克的任务

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：

\[dp[i] = \begin{cases} dp[i - 1] + 1 & \texttt{don't have work}\\ \max \{dp[i + T[s]\} & \texttt{have work at } \color{red}{\texttt{s}} \end{cases} \]

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：

\[dp[i] = \begin{cases} dp[i + 1] + 1 & \texttt{don't have work}\\ \max \{dp[i + T[s]\} & \texttt{have work at } \color{red}{\texttt{s}} \end{cases} \]

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\n", dp[1]);
    return 0;
}