算法分析第十次作业

1. 问题

   给出n个节目，给出每个节目开始时间和结束时间，同一时间段只能播放一个节目，问最多可以播放多少节目

2. 解析

  贪心策略是按结束时间从小到大排序，贪心选择和最后一个节目相容的节目，最后得到就是的答案

3. 设计

   排序方式

struct node {
    int l,r;
    bool operator <(const node &a){
        return r<a.r;
    }
}t[maxn];

选择方式

sort(t+1,t+n+1);
puts("选择:");
int l=0;
for(int i=1;i<=n;i++){
    if(t[i].l>=l) printf("%d %d
",t[i].l,t[i].r),l=t[i].r;
}

4. 分析

  O(n)

5. 源码

https://github.com/Tinkerllt/algorithm-work.git

#include<stdio.h>
#include<stdio.h>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<set>
#include<deque>
#include<queue>
#include<vector>
//#include<unordered_map>
#include<map>
#include<stack>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define Pii pair<ll,int>
#define m_p make_pair
#define l_b lower_bound
#define u_b upper_bound
const int inf = 0x3f3f3f3f;
const ll linf = 0x3f3f3f3f3f3f3f3f;
const int maxn = 3e5 + 11;
const int maxm = 600 + 11;
const int mod = 1e9 + 7;
const double eps = 1e-5;
inline ll rd() { ll x = 0, f = 1; char ch = getchar(); while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }while (ch >= '0'&&ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }return x * f; }
inline ll qpow(ll a, ll b, ll p) { ll res = 1; while (b) { if (b & 1) { res *= a; res %= p; }b >>= 1; a = a * a%p; }return res; }
inline ll gcd(ll a, ll b) { if (b == 0) return a; return gcd(b, a%b); }
//iterator
//head
//priority_queue
struct node {
    int l,r;
    bool operator <(const node &a){
        return r<a.r;
    }
}t[maxn];
int main(){
    int n=rd();
    for(int i=1;i<=n;i++){
        t[i].l=rd(),t[i].r=rd();
    } 
    sort(t+1,t+n+1);
    puts("选择:");
    int l=0;
    for(int i=1;i<=n;i++){
        if(t[i].l>=l) printf("%d %d
",t[i].l,t[i].r),l=t[i].r;
    }
}