CODEVS 1214

Posted 2020-09-29 %%%%%

1214 线段覆盖

题意：中文题

思路：贪心，先按l排序后按r排序，每次标记当前剩下线段的最右端，因为l是从小到达有序的，所以不需要考虑左端点，只需要考虑右边尽量往左端点取就是了

AC代码：

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#define ll long long
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a) memset(a,0,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;
const ll mod=1e9+7;

struct Node{
    int l, r;
    bool friend operator< ( Node a, Node b){
        if(a.l==b.l) return a.r<b.r;
        return a.l<b.l;
    }
};
Node arr[105];
int n,ans=-1;
int main(){
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n;
    for(int i=1; i<=n; ++i){
        int x,y;
        cin>>x>>y;
        arr[i].l=min(x,y);
        arr[i].r=max(x,y);
    }
    sort(arr+1,arr+1+n);
    int p=inf;
    for(int i=1; i<=n; ++i){
        if(arr[i].r>=p){
            if(arr[i].l<p){
                ans++;
            }
            else p=arr[i].r;
        }
        else{
            ans++;
            p=arr[i].r;
        }
    }
    cout<<n-ans<<"\n";
    return 0;
}