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  • Intervals ZOJ

    Chiaki has n intervals and the i-th of them is [liri]. She wants to delete some intervals so that there does not exist three intervals ab and c such that aintersects with bb 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.

    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 li and ri (1 ≤ li < ri ≤ 109) denoting the i-th interval. Note that for every 1 ≤ i < j ≤ nli ≠ lj or ri ≠ rj.

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

    <h4< dd="">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.

    <h4< dd="">Sample Input

    1
    11
    2 5
    4 7
    3 9
    6 11
    1 12
    10 15
    8 17
    13 18
    16 20
    14 21
    19 22
    

    <h4< dd="">Sample Output

    4
    3 5 7 10

    删除最少的区间使得不存在 3个区间两两重合

    #include <cstdio>
    #include <cstring>
    #include <queue>
    #include <cmath>
    #include <algorithm>
    #include <set>
    #include <iostream>
    #include <map>
    #include <stack>
    #include <string>
    #include <vector>
    #define  pi acos(-1.0)
    #define  eps 1e-6
    #define  fi first
    #define  se second
    #define  lson l,m,rt<<1
    #define  rson m+1,r,rt<<1|1
    #define  bug         printf("******
    ")
    #define  mem(a,b)    memset(a,b,sizeof(a))
    #define  fuck(x)     cout<<"["<<x<<"]"<<endl
    #define  f(a)        a*a
    #define  sf(n)       scanf("%d", &n)
    #define  sff(a,b)    scanf("%d %d", &a, &b)
    #define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
    #define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
    #define  pf          printf
    #define  FRE(i,a,b)  for(i = a; i <= b; i++)
    #define  FREE(i,a,b) for(i = a; i >= b; i--)
    #define  FRL(i,a,b)  for(i = a; i < b; i++)
    #define  FRLL(i,a,b) for(i = a; i > b; i--)
    #define  FIN         freopen("DATA.txt","r",stdin)
    #define  gcd(a,b)    __gcd(a,b)
    #define  lowbit(x)   x&-x
    #pragma  comment (linker,"/STACK:102400000,102400000")
    using namespace std;
    typedef long long  LL;
    typedef unsigned long long ULL;
    const int INF = 0x7fffffff;
    const int mod = 1e9 + 7;
    const int maxn = 5e5 + 10;
    int t, n;
    struct node {
        int x, y, id;
    } p[maxn];
    int cmp1(node a, node b) {
        if (a.x == b.x) return a.y < b.y;
        return a.x < b.x;
    }
    int cmp2(node a, node b) {
        if (a.y == b.y) return a.x < b.x;
        return a.y > b.y;
    }
    int check(node a, node b, node c) {
        int cnt1 = (b.x <= a.y);
        int cnt2 = (c.x <= b.y && c.x <= a.y);
        return (cnt1 && cnt2);
    }
    int ans[maxn];
    int main() {
        sf(t);
        while(t--) {
            sf(n);
            for(int i = 1 ; i <= n ; i++) {
                sff(p[i].x, p[i].y);
                p[i].id = i;
            }
            sort(p + 1, p + 1 + n, cmp1);
            node x[5];
            x[0] = p[1], x[1] = p[2];
            int num = 0;
            for (int i = 3 ; i <= n ; i++) {
                x[2] = p[i];
                sort(x, x + 3, cmp1);
                int flag = check(x[0], x[1], x[2]);
                sort(x, x + 3, cmp2);
                if (flag) {
                    ans[num++] = x[0].id;
                    swap(x[0], x[2]);
                }
            }
            printf("%d
    ", num);
            if (!num) {
                printf(" 
    ");
                continue;
            }
            sort(ans, ans + num);
            if (!num) {
                printf(" 
    ");
                continue;
            }
            for (int i = 0 ; i < num ; i++) printf("%d ", ans[i]);
            printf("
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/qldabiaoge/p/9544785.html
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