HDU 5124 lines

最新推荐文章于 2019-01-21 19:52:48 发布

原创最新推荐文章于 2019-01-21 19:52:48 发布 · 340 阅读

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lines

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1439 Accepted Submission(s): 594

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

Sample Output

3
1

Source

BestCoder Round #20

有两种方法：

第一种是直接把端点排序，起点的值为1，终点的值为-1，然后过一遍取最大值。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
#include<vector>
#include<map>
#include<cmath>
#include<fstream>
#include<stack>
using namespace std;
typedef long long ll;
struct node
{
    int x, v;
}s[200020];
int cmp(node a, node b)
{
    if(a.x == b.x)
        return a.v>b.v;
    return a.x <b.x;
}
int main()
{
   int t, n;
   cin>>t;
   while(t--)
   {
       scanf("%d", &n);
       for(int i = 0; i<n; i++)
       {
           scanf("%d", &s[i<<1].x);
           s[i<<1].v = 1;
           scanf("%d", &s[i<<1|1].x);
           s[i<<1|1].v = -1;
       }
       sort(s, s+2*n, cmp);
       int cnt = 0, ans = 0;
       for(int i = 0; i<2*n; i++)
       {
           cnt += s[i].v;
           ans = max(cnt , ans);
       }
       cout<<ans<<endl;
   }
}

第二种:先离散化，然后线段树区间更新，取最大值。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
#include<vector>
#include<map>
#include<cmath>
#include<fstream>
using namespace std;
struct node
{
    int l,r, maxx, mark;
} s[200200<<2];
int x[200020], k, a[100010], b[100010];
void pushup(int cnt)
{
    s[cnt].maxx = max(s[cnt<<1].maxx , s[cnt<<1|1].maxx);
}
void pushdown(int cnt)
{
    if(s[cnt].mark)
    {
        //int l = len - (len/2);
        s[cnt<<1].mark += s[cnt].mark ;
        s[cnt<<1|1].mark += s[cnt].mark;
        s[cnt<<1].maxx += s[cnt].mark;
        s[cnt<<1|1].maxx += s[cnt].mark;
        s[cnt].mark = 0;
    }
}
void build_tree(int l, int r, int cnt)
{
    s[cnt].l = l;
    s[cnt].r = r;
    s[cnt].maxx = 0;
    s[cnt].mark = 0;
    if(l == r)
        return ;
        int m= (l+r)>>1;
    build_tree(l, m, cnt<<1);
    build_tree(m+1, r, cnt<<1|1);
}
void update(int l, int r, int cnt)
{
    if(s[cnt].l >= l && s[cnt].r <= r)
    {
        s[cnt].mark += 1;
        s[cnt].maxx += 1;
        return ;
    }
    pushdown(cnt);
    int m = (s[cnt].l + s[cnt].r)>>1;
    if(l <= m)
        update(l, r, cnt<<1);
    if(r > m)
        update(l, r, cnt<<1|1);
    pushup(cnt);
}
int main()
{
    int t, n, u, v;
    cin>>t;
    while(t--)
    {
        scanf("%d", &n);
        k = 0;
        for(int i = 1; i<=n; i++)
        {
            scanf("%d%d", &a[i], &b[i]);
            x[k++] = a[i];
            x[k++] = b[i];
        }
        sort(x,x+2*n);
        //for(int i = 0; i<2*n; i++)
      //      printf("%d ", x[i]);
      //  printf("\n");
        build_tree(1, 2*n, 1);
        for(int i = 1; i<=n; i++)
        {
            int l = lower_bound(x, x+2*n, a[i]) - x + 1;
            int r = lower_bound(x, x+2*n, b[i]) - x + 1;
          //  printf("%d %d %d %d\n", l, r, a[i], b[i]);
            update(l, r, 1);
        }
        printf("%d\n", s[1].maxx);
    }
    return 0;
}