扫描线（线段树）+贪心 ZOJ 3953

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5572

Intervals

Time Limit: 1 Second Memory Limit: 65536 KB Special Judge

Chiaki has n intervals and the i-th of them is [l_i, r_i]. 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 l_a ≤ x ≤ r_a and l_b ≤ x ≤ r_b.

Input

There 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 l_i and r_i (1 ≤ l_i < r_i ≤ 10⁹) denoting the i-th interval. Note that for every 1 ≤ i < j ≤ n, l_i ≠ l_j or r_i ≠ r_j.

It is guaranteed that the sum of all n does not exceed 500000.

Output

For 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 Input

Sample Output

4

3 5 7 10

Author: LIN, Xi
Source: The 17th Zhejiang University Programming Contest Sponsored by TuSimple

题目大意：给你n个区间，这n个区间里面可能存在这样的区间，即a交b，b交c，c交a。问最少删除多少个，才能使得不存在这样的区间，并输出删除的是什么区间。

思路：扫面线一遍，从左往右扫（就是单点查询，md我今天才知道原来这是一维扫描线，打比赛的时候完全不知道）。然后每次都往priority_queue里面放入目前询问到节点的右端点，当cnt>=3的时候贪心的删除最右边的端点即可。

//看看会不会爆int!数组会不会少了一维！

//取物问题一定要小心先手胜利的条件

#include <bits/stdc++.h>

using namespace std;

#pragma comment(linker,"/STACK:102400000,102400000")

#define LL long long

#define ALL(a) a.begin(), a.end()

#define pb push_back

#define mk make_pair

#define fi first

#define se second

#define haha printf("haha\n")

const int maxn =  + ;

int n;

vector<int> ve;

int l[maxn], r[maxn];

vector<pair<int, int> > qujian[maxn];

struct Point{

    int val, lazy;

}tree[maxn << ];

void buildtree(int l, int r, int o){

    tree[o].val = tree[o].lazy = ;

    if (l == r){

        return ;

    }

    int mid = (l + r) / ;

    if (l <= mid) buildtree(l, mid, o << );

    if (r > mid) buildtree(mid + , r, o <<  | );

}

void push_down(int o){

    int lb = o << , rb = o <<  | ;

    tree[lb].lazy += tree[o].lazy; tree[lb].val += tree[o].lazy;

    tree[rb].lazy += tree[o].lazy; tree[rb].val += tree[o].lazy;

    tree[o].lazy = ;

}

void update(int ql, int qr, int l, int r, int o, int add){

    //printf("ql = %d qr = %d l = %d r = %d\n", ql, qr, l, r);

    if (ql <= l && qr >= r){

        tree[o].lazy += add;

        tree[o].val += add;

        return ;

    }

    if (tree[o].lazy != ) push_down(o);

    int mid = (l + r) / ;

    if (ql <= mid) update(ql, qr, l, mid, o << , add);

    if (qr > mid) update(ql, qr, mid + , r, o <<  | , add);

    tree[o].val = max(tree[o << ].val, tree[o <<  | ].val);

}

int query(int ql, int qr, int l, int r, int o){

    if (ql <= l && qr >= r){

        return tree[o].val;

    }

    if (tree[o].lazy != ) push_down(o);

    int mid = (l + r) / ;

    int maxval = ;

    if (ql <= mid) maxval = max(maxval, query(ql, qr, l, mid, o <<));

    if (qr > mid) maxval = max(maxval, query(ql, qr, mid + , r, o << | ));

    tree[o].val = max(tree[o << ].val, tree[o <<  | ].val);

    return maxval;

}

void solve(){

    int ans = ;

    vector<int> tmp;

    buildtree(,  * n, );

    for (int i = ; i <= n; i++){

        update(l[i], r[i], ,  * n, , );

    }

    priority_queue<pair<int, int> > que;

    for (int i = ; i <=  * n; i++){

        if (qujian[i].size()){

            for (int j = ; j < qujian[i].size(); j++){

                que.push(qujian[i][j]);

            }

            int cnt = query(i, i, ,  * n, );

            while (cnt >= ){

                cnt--;

                pair<int, int> pa = que.top(); que.pop();

                update(i, pa.fi, , *n, , -);

                ans++;

                tmp.push_back(pa.se);

            }

            /*for (int j = qujian[i].size() - 1; j >= 0; j--){

                int L = i, R = qujian[i][j].fi;

                int cnt = query(L, R, 1, 2 * n, 1);

                 cout<<L << ' ' << R<< ' ' <<cnt<<endl;

                if (cnt >= 3){

                    ans++;

                    update(L, R, 1, 2 * n, 1, -1);

                    tmp.push_back(qujian[i][j].se);

                }

                else break;

            }*/

        }

    }

    sort(ALL(tmp));

    printf("%d\n", ans);

    for (int i = ; i < tmp.size(); i++){

        printf("%d%c", tmp[i], i == tmp.size()- ? '\n' : ' ');

    }

}

int main(){

    int t; cin >> t;

    while (t--){

        ve.clear();

        scanf("%d", &n);

        for (int i = ; i <=  * n; i++) qujian[i].clear();

        for (int i = ; i <= n; i++){

            scanf("%d%d", l + i, r + i);

            ve.push_back(l[i]);

            ve.push_back(r[i]);

        }

        sort(ALL(ve));

        ve.erase(unique(ALL(ve)), ve.end());

        for (int i = ; i <= n; i++){

            l[i] = lower_bound(ALL(ve), l[i]) - ve.begin() + ;

            r[i] = lower_bound(ALL(ve), r[i]) - ve.begin() + ;

            qujian[l[i]].push_back(mk(r[i], i));

        }

        for (int i = ; i <=  * n; i++){

            if (qujian[i].size()){

                sort(ALL(qujian[i]));

            }

        }

        solve();

    }

    return ;

}