hdoj5124lines【贪心 or 离散化】

lines

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1364    Accepted Submission(s): 567

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

251 2 2 22 43 45 100051 12 23 34 45 5

 

Sample Output

31

 

Source

BestCoder Round #20

 

题意：给出一些线段求一个被线段覆盖最多的点的覆盖线段数

可以用离散化做但离散化还不知道怎么做此题也可以用贪心做

贪心思路是把一条线段拆成两个点后排序，那么此时当前位置如果遇到一个起点，那么被覆盖的线段数要+1，遇到终点就是一条线段完结了被覆盖的数量-1，找出过程中的最大值即可。

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>using namespace std;const int maxn=100010;int s[maxn];int e[maxn];bool cmp(int a,int b){    return a<b;}int MAX(int a,int b){    return a>b?a:b;}int main(){    int t,i,j,k,n;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        for(i=0;i<n;++i){            scanf("%d%d",&s[i],&e[i]);        }        sort(s,s+n,cmp);        sort(e,e+n,cmp);        int ans=0,temp=0,x=0,y=0;        while(x<n){            if(s[x]<=e[y]){                temp++;x++;            }            else {                temp--;y++;            }            ans=MAX(ans,temp);        }        printf("%d\n",ans);    }    return 0;}