HDU 5124 lines

lines

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1413    Accepted Submission(s): 576

Problem Description John has several lines. The lines are covered on the X axis. Let A is a point which is covered by the most lines. John wants to know how many lines cover A.  

Input The first line contains a single integer  T(1≤T≤100) (the data for  N>100  less than 11 cases),indicating the number of test cases.

Each test case begins with an integer  N(1≤N≤105) ,indicating the number of lines.

Next N lines contains two integers  Xi  and  Yi(1≤Xi≤Yi≤109) ,describing a line.  

Output For each case, output an integer means how many lines cover A.  

Sample Input

2
5
1 2 
2 2
2 4
3 4
5 1000
5
1 1
2 2
3 3
4 4
5 5
        

Sample Output

3

   
1

   


          
//樹狀數組加離散化
//暑假學的樹狀數組和資料離散化又忘得差不多了。。。重新學了一遍。。。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

typedef long long ll;
const int N = 100100;
int bit[N*2], a[N*2];
int n;
struct node
{
    int v, u;
}s[N];

int sum(int i)
{
    int s = 0;
    while(i > 0)
    {
        s += bit[i];
        i -= i & - i;
    }
    return s;
}
void add(int i, int x)
{
    while(i <= 2 * n) //離散化後最多n*2個點
    {
        bit[i] += x;
        i += i & - i;
    }
}

int main()
{
    int t;

    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        memset(bit, 0, sizeof bit);

        int k = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d", &s[i].v, &s[i].u);
            a[k++] = s[i].v;
            a[k++] = s[i].u;
        }

        sort(a, a + k);
        for(int i = 0; i < n; i++) //離散化
        {
            s[i].v = lower_bound(a, a + k, s[i].v) - a + 1;
            s[i].u = lower_bound(a, a + k, s[i].u) - a + 1;
        }

        for(int i = 0; i < n; i++)
        {
            add(s[i].v, 1);
            add(s[i].u + 1, -1);
        }

        int res = -1;
        for(int i = 1; i <= 2 * n; i++) //離散化後最多n*2個點
            res = max(res, sum(i));

        printf("%d\n", res);
    }

    return 0;
}
           
        
 
線段樹的做法，用cnt記錄被覆寫的次數，不過效率較慢，所用時間接近限制，有一半的機率TLE，看人品。。。          
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;

const int N = 100100;
int a[N], b[N], tmp[N*2];
struct node
{
    int l, r, cnt, mark;
}s[N*8];

void PushUp(int k)
{
    s[k].cnt = max(s[k<<1].cnt, s[k<<1|1].cnt);
}

void PushDown(int k)
{
    if(s[k].mark)
    {
        s[k<<1].cnt += s[k].mark;
        s[k<<1|1].cnt += s[k].mark;
        s[k<<1].mark += s[k].mark;
        s[k<<1|1].mark += s[k].mark;
        s[k].mark = 0;
    }
}

void init(int l, int r, int k)
{
    s[k].l = l, s[k].r = r, s[k].cnt = s[k].mark = 0;
    if(l == r)
        return;
    int mid = (l + r) >> 1;
    init(l, mid, k << 1);
    init(mid + 1, r, k << 1|1);
}

void update(int l, int r, int k)
{
    if(l <= s[k].l && s[k].r <= r)
    {
        s[k].cnt++;
        s[k].mark++;
        return;
    }

    PushDown(k);
    int mid = (s[k].l + s[k].r) >> 1;
    if(l <= mid) update(l, r, k << 1);
    if(r > mid) update(l, r, k << 1|1);
    PushUp(k);
}

int main()
{
    int t, n;

    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);

        int k = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d", a + i, b + i);
            tmp[k++] = a[i];
            tmp[k++] = b[i];
        }

        sort(tmp, tmp + k);
        int max1 = -1;
        for(int i = 0; i < n; i++)
        {
            a[i] = lower_bound(tmp, tmp + k, a[i]) - tmp + 1;
            b[i] = lower_bound(tmp, tmp + k, b[i]) - tmp + 1;
            max1 = max(max1, max(a[i], b[i]));
        }

        init(1, max1, 1);
        for(int i = 0; i < n; i++)
            update(a[i], b[i], 1);

        printf("%d\n", s[1].cnt);
    }

    return 0;
}