ZOJ3953 Intervals 【贪心 双线程活动分配问题】

IntervalsTime Limit: 1 Second      Memory Limit: 65536 KB      Special JudgeChiaki has n intervals and the i-th of them is [li, ri]. She wants to delete some intervals so that there does not exist three intervals a, b and c such that a intersects with b, b intersects with c and c intersects with a.Chiaki is interested in the minimum number of intervals which need to be deleted.Note that interval a intersects with interval b if there exists a real number x such that la ≤ x ≤ ra and lb ≤ x ≤ rb.InputThere are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:The first line contains an integer n (1 ≤ n ≤ 50000) -- the number of intervals.Each of the following n lines contains two integers li and ri (1 ≤ li < ri ≤ 109) denoting the i-th interval. Note that for every 1 ≤ i < j ≤ n, li ≠ lj or ri ≠ rj.It is guaranteed that the sum of all n does not exceed 500000.OutputFor each test case, output an integer m denoting the minimum number of deletions. Then in the next line, output m integers in increasing order denoting the index of the intervals to be deleted. If m equals to 0, you should output an empty line in the second line.Sample Input1112 54 73 96 111 1210 158 1713 1816 2014 2119 22Sample Output43 5 7 10Author: LIN, XiSource: The 17th Zhejiang University Programming Contest Sponsored by TuSimple

明显，问题就是求：删除至少多少个区间，使得不存在任意一点x被>=3个区间覆盖
先简化一下问题
假如问题改为，不存在x被>=2个区间覆盖
那直接贪心，求一次活动分配问题就行了，标记哪个区间被用掉了，没被用的全部是被删掉的

在已经按以上的方法 ，选择了一部分区间后
显然 ，再选择任意一个还没被选择的区间，必然会让某一点x，被2个区间覆盖
那再对剩余的区间求一次活动分配
这样 如果再选择剩余的任意区间，都会使得某一点x被3个区间覆盖
然而 如果遇到
[1,2] [1,3] [4,100] [3,101] 这样的情况
经典的活动分配问题会选出
[1,2] [4,100]
[1,3]
最后删掉[3,101]

而显然
[1,2] [3,101]
[1,3] [4,100]更优

那就修改一下算法
同时进行2个线程的活动分配
假如线程1的最晚活动结束时间为t1
线程2的最晚结束时间为t2

对一个待选择的活动，优先添加到t更大的线程后面(前提是该活动开始时间>t)

#include<iostream>#include<cstdlib>#include<cstdio>#include<string>#include<vector>#include<deque>#include<queue>#include<algorithm>#include<set>#include<map>#include<stack>#include<ctime>#include<string.h>#include<math.h>#include<list>using namespace std;#define ll long long#define pii pair<int,int>const int inf = 1e9 + 7;const int N = 50000 + 5;struct S{    int l,r;    int idx;}a[N];bool used[N];bool cmp(const S&a,const S&b){    if(a.r!=b.r){        return a.r<b.r;//结束时间升序    }    return a.l>b.l;//开始时间递减序}vector<int>ans;void slove(int n){    sort(a,a+n,cmp);    fill(used,used+n+1,0);    int end1=0,end2=0;//线程1,2的最晚结束时间    for(int i=0;i<n;++i){        if(used[i]==0){            int*enda=&end1;            int*endb=&end2;            if(end2>end1){                swap(enda,endb);            }            if(*enda<a[i].l){                *enda=a[i].r;                used[i]=1;            }            else{                if(*endb<a[i].l){                    *endb=a[i].r;                    used[i]=1;                }            }        }    }    ans.clear();    for(int i=0;i<n;++i){//被删除的点        if(used[i]==0){            ans.push_back(a[i].idx);        }    }    sort(ans.begin(),ans.end());    printf("%d\n",ans.size());    if(ans.empty()){        putchar('\n');    }    else{        printf("%d",ans[0]);        for(int i=1;i<ans.size();++i){            printf(" %d",ans[i]);        }        putchar('\n');    }}int main(){    //freopen("/home/lu/Documents/r.txt","r",stdin);    //freopen("/home/lu/Documents/w.txt","w",stdout);    int T;    scanf("%d",&T);    while(T--){        int n;        scanf("%d",&n);        for(int i=0;i<n;++i){            scanf("%d%d",&a[i].l,&a[i].r);            a[i].idx=i+1;        }        slove(n);    }    return 0;}