HDU 5124 lines
2014-11-30 23:56
375 查看
lines
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 515 Accepted Submission(s): 241
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each case, output an integer means how many lines cover A.
[align=left]Sample Input[/align]
2
5
1 2
2 2
2 4
3 4
5 1000
5
1 1
2 2
3 3
4 4
5 5
[align=left]Sample Output[/align]
3
1
BC#20的1002题
区间的数据范围去到1e8
只要离散一下再用差分标记搞就可以了
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <cmath> #include <vector> #include <queue> #include <map> #include <set> #include <stack> #include <algorithm> using namespace std; typedef long long LL; typedef pair<int,int> pii; #define X first #define Y second const int N = 300010 ; const int inf = 1e9+7; int cnt , tot ; struct node { int x , xx , id ; }e ; bool cmp1( const node &a , const node &b ){ return a.x < b.x ; } bool cmp2( const node &a , const node &b ){ return a.id < b.id ; } void run() { memset( cnt , 0 , sizeof cnt ) ; int n ; tot = 0 ; scanf("%d",&n); for( int i = 0 ; i < 2 * n ; ++i ){ scanf("%d",&e[i].x); e[i].id = i ; } sort( e , e + 2 * n , cmp1 ); e[0].xx = 1 ; for( int i = 1 ; i < 2 * n ; ++i ){ e[i].xx = ( e[i].x == e[i-1].x ? e[i-1].xx : e[i-1].xx + 1 ); } sort( e, e + 2 * n , cmp2 ) ; for( int i = 0 ; i < 2 * n ; i += 2 ){ cnt[ e[i].xx ] ++; cnt[ e[i+1].xx + 1 ] --; } int ans = 0 , sum = 0 ; for( int i = 0 ; i <= 2 * n ; ++i ){ sum += cnt[i]; ans = max( ans , sum ) ; } printf("%d\n",ans); } int main() { #ifdef LOCAL freopen("in.txt","r",stdin); #endif // LOCAL ios::sync_with_stdio(false); int _ ; scanf("%d",&_); while( _-- ) run() ; }
View Code
相关文章推荐
- hdu 5124 lines(树状数组)
- hdu5124——lines
- HDU 5124 lines
- HDU 5124 lines
- [思路题] hdu 5124 lines
- HDU - 5124 lines
- HDU 5124 lines 最多区间覆盖
- hdu 5124——lines
- BestCoder20 1002.lines (hdu 5124) 解题报告
- HDU 5124 lines
- 【HDU - 5124】 lines 【树状数组+离散化】
- 【扫描线】HDU 5124 lines
- hdu 5124 lines(离散化)
- 哈理工练习赛 HDU 5124 lines
- hdu 5124 BestCoder #20 1002 lines 解题报告
- hdu 5124 lines (线段树+离散化)
- HDU 5124 lines 最大区间重叠点(离散化)
- 【瞎搞】HDU 5124 lines
- hdu 5124 lines
- hdu 5124 lines(Bestcoder Round #20)