bzoj 3410 [Usaco2009 Dec]Selfish Grazing 自私的食草者 贪心

题面

题目传送门

解法

特别基础的贪心题

就是问最多能选出多少个区间，使得这些区间两两没有交集

直接按照右端点排序，然后贪心取就可以了

时间复杂度：\(O(n\ log\ n)\)

代码

#include <bits/stdc++.h>
#define N 50010
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
    x = 0; int f = 1; char c = getchar();
    while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Node {
    int l, r;
    bool operator < (const Node &a) const {
        return r < a.r;
    }
} a[N];
int main() {
    int n; read(n);
    for (int i = 1; i <= n; i++)
        read(a[i].l), read(a[i].r), a[i].r--;
    sort(a + 1, a + n + 1);
    int ans = 0, lr = 0;
    for (int i = 1; i <= n; i++)
        if (a[i].l > lr) ans++, lr = a[i].r;
    cout << ans << "\n";
    return 0;
}