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HDU 5124 lines 最多区间覆盖
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Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 453 Accepted Submission(s): 220
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 integerN(1≤N≤105) ,indicating the number of lines.
Next N lines contains two integersXi and Yi(1≤Xi≤Yi≤109) ,describing a line.
Each test case begins with an integer
Next N lines contains two integers
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
Source
BestCoder Round #20
官方题解:
我们可以将一条线段[xi,yi] 分为两个端点xi 和(yi)+1 ,在xi 时该点会新加入一条线段,同样的,在(yi)+1 时该点会减少一条线段,因此对于2n个端点进行排序,令xi 为价值1,yi 为价值-1,问题转化成了最大区间和,因为1一定在-1之前,因此问题变成最大前缀和,我们寻找最大值就是答案,另外的,这题可以用离散化后线段树来做。复杂度为排序的复杂度即nlgn ,另外如果用第一种做法数组应是2n,而不是n,由于各种非确定性因素我在小数据就已经设了n=10W的点。
//953MS 1792K #include<stdio.h> #include<algorithm> using namespace std; pair<int,int>p[200007]; int main() { int t; scanf("%d",&t); while(t--) { int n,a,b; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d%d",&a,&b); p[i*2]=make_pair(a,1); p[i*2+1]=make_pair(b+1,-1); } sort(p,p+2*n);//对坐标进行排序 int ans=0,maxx=0; for(int i=0;i<2*n;i++) { ans+=p[i].second; if(ans>maxx)maxx=ans; } printf("%d\n",maxx); } return 0; }
HDU 5124 lines 最多区间覆盖
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