HDU - 5124 lines

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

 

　　真是个好题，通过这个题我学会了离散化的一点知识，首先把所有的坐标映射到x轴上，然后将坐标压缩。怎么压缩呢？

可以用一个v数组，直接存贮位置就可以了，那么这样就完成了坐标的压缩。

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <map>
#include <vector>

using namespace std;

#define MAXN 200005

typedef pair<int,int> PII;

int x[MAXN];
PII p[MAXN];
int v[MAXN];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        int cnt = 0;
        for(int i = 0;i<n;i++){
            scanf("%d %d",&p[i].first,&p[i].second);
            x[cnt++] = p[i].first;
            x[cnt++] = p[i].second;
        }
        sort(x,x+cnt);
        cnt = unique(x,x+cnt)-x;
        for(int i = 0;i<cnt;i++){
            v[i] = 0;
        }
        int sum = 0;
        for(int i = 0;i<n;i++){
            int l = lower_bound(x,x+cnt,p[i].first)-x;
            int r = lower_bound(x,x+cnt,p[i].second)-x;
            v[l]++;
            v[r+1]--;
        }
        int s = 0;
        int mx = 0;
        for(int i = 0;i<cnt;i++){
            s+=v[i];
            mx = max(mx,s);
        }
        printf("%d\n",mx);
    }
    return 0;
}

 

　　当然不一定非得把坐标映射到x轴上，vector <pair<int,int> > v; v[i].first 用来把坐标排序，起到了映射的作用。然后数组的下标就起到了压缩的作用。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

vector <pair<int,int> > v;

int main () {
    int T;
    cin >> T;
    while (T--) {
        v.clear();
        int N;
        scanf("%d",&N);
        for (int i = 0;i < N;i++) {
            int x;
            scanf("%d",&x);
            v.push_back(make_pair(x,1));
            scanf("%d",&x);
            v.push_back(make_pair(x + 1,-1));
        }
        sort(v.begin(), v.end());
        int ans = 0;
        int maxn = 0;
        for (int i = 0;i < v.size();i++) {
            ans += v[i].second;
            maxn = max(maxn,ans);
        }
        cout << maxn << endl;
    }
}