cf 1249 D2. Too Many Segments (hard version) （贪心+multiset）

Meaning of the questions:

There are n line segments, each segment covering points [L _I , R & lt _I ], if the points are more than k line segments covering, then the point of dead pixels,

Q. Which at least a few segments removed, can make no dead pixels?

 

Ideas:

Point sweep from left to right, if the point of dead pixels, then remove some of the points covered in this segment, from r _i s largest delete start, no need to consider L _i is because you have left to ensure that no dead pixels, then delete out of the line of course R & lt _I better.

Multiset line with current maintenance coverage at this point is which a few.

I know today, sort when the externally defined function cmp structure may be covered in the overloaded operators '<'.

It is not really a big brother less stl, when the test sample RE, the two cross-hair is also RE, long tone.

 

Code:

#include <stdio.h>
#include <string.h>
#include <cmath>
#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
typedef long long int ll;
const double pi = acos(-1);
const int maxn = 2e5 + 10;
const int inf = 0x3f3f3f3f;
struct node
{
    int l,r,id;
    friend bool operator <(node x,node y){
        return x.r < y.r;
    }
}a[maxn];

bool cmp(node x,node y)
{
    return x.l < y.l;
}
int vis[maxn];
multiset<node>st;
int main()
{
    multiset<node>::iterator it;
    int n,k,ans,mx;
    while(scanf("%d%d",&n,&k) != EOF){
        ans = 0;
        mx = 0;
        st.clear();
        memset(vis,0,sizeof(vis));
        for(int i = 1;i <= n;i++){
            scanf("%d%d",&a[i].l,&a[i].r);
            mx = max(mx,a[i].r);
            a[i].id = i;
        }
        sort(a + 1,a + n + 1,cmp);
        int cnt = 1,now = 0;
        for(int i = 1;i <= mx;i++){
            it = st.begin();
            while(it != st.end() && (*it).r < i){
                st.erase(it);
                it = st.begin();
                now--;
            }
            while(cnt <= n && a[cnt].l <= i){
                st.insert(a[cnt]);
                cnt++;
                now++;
            }
            it = st.end();
            while(now > k){
                it--;
                vis[(*it).id] = 1;
                st.erase(it);
                
                it = st.end();
                now--;
                ans++;
            }
        }
        printf("%d\n",ans);
        for(int i = 1;i <= n;i++){
            if(vis[i])
                printf("%d ",i);
        }
        puts("");
    }
    return 0;
}

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