P1868 饥饿的奶牛

Problem

给定\(n\)个闭区间\([x_i,y_i]\)，要选若干个区间，使其区间长度和最大，且无交集。
\(n \le 1.5 \times 10^5,1 \le x_i,y_i \le 3 \times 10^6\)

Solution

设\(m = \max \{y_i\}\)。
设\(dp_i\)为\([i,n]\)中的最大数,\(g_i\)为以\(i\)为左端点的区间的序号,\(len_i\)表示第\(i\)个区间的长度。则

\[dp_i = \begin{cases} dp_{i + 1} & g_i = \emptyset \\ \max (dp_{i + 1},\max_{j \in g_i} \{dp_{i + len_j} + len_j\} ) & g_i \neq \emptyset \end{cases} \]

# include <bits/stdc++.h>
using namespace std;
const int N = 1500005,M = 3e6 + 5;
int n;
pair<int,int> a[N];
vector <int> g[N];
int dp[M];
int Left[M],Right[M],len[N];
int m;
int main(void)
{
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d%d",&a[i].first,&a[i].second);
        g[a[i].first].push_back(i);
        len[i] = a[i].second - a[i].first + 1;
        m = max(m,a[i].second);
    }
    for(int i = m; i >= 1; i--)
    {
        dp[i] = dp[i + 1];
        if(g[i].size())
        {
            for(int j = 0; j < (int)g[i].size(); j++)
            {
                dp[i] = max(dp[i],dp[i + len[g[i][j]]] + len[g[i][j]]);
            }
        }
    }
    printf("%d\n",dp[1]);
    return 0;
}

标签：le,P1868,max,len,int,饥饿,区间,奶牛,dp
来源： https://www.cnblogs.com/luyiming123blog/p/14999588.html