【C++竞赛 G】Lines

Time Limit: 3s Memory Limit: 64MB
问题描述
Ljr has several lines. The lines are covered on the X axis. Let A is a point which is covered by the most lines. Ljr wants to know how many lines cover A.
输入描述
The first line contains a single integer T(1≤T≤100) (the data for N>100 less than 10 cases), indicating the number of test cases. Each test case begins with an integerN(1≤N≤〖10〗^5), indicating the number of lines. Next N lines contains two integers X_i and Y_i (-〖10〗^9≤X_i,Y_i≤〖10〗^9), describing a line.
输出描述
For each case, output an integer means how many lines cover A.
输入样例
2
5
1 2
2 3
2 4
3 4
5 1000
5
1 2
3 4
5 6
7 8
9 10
输出样例
3
1
 
【题目链接】:

【题解】

把区间端点离散化一下，然后就转换成区间最大值的问题了；
写个线段树就好;
x<=y不一定成立;

【完整代码】

#include <bits/stdc++.h>#define rep1(i,a,b) for (int i = a;i <= b;i++)using namespace std;#define pb push_back;const int MAXN = 1e5+10;int ma[MAXN*2*4],tag[MAXN*2*4];struct abc{    int l,r;};abc aa[MAXN];vector <int> a;map <int,int> dic;void push_down(int rt){    tag[rt<<1]+=tag[rt];    tag[rt<<1|1]+=tag[rt];    ma[rt<<1]+=tag[rt];    ma[rt<<1|1]+=tag[rt];    tag[rt] = 0;}void up_data(int L,int R,int l,int r,int rt){    //printf("%d %d\n",l,r);    if (L<=l && r <= R)    {        ma[rt]++;        tag[rt]++;        return;    }    if (tag[rt]!=0)        push_down(rt);    int m = (l+r)>>1;    if (L<=m)        up_data(L,R,l,m,rt<<1);    if (m<R)        up_data(L,R,m+1,r,rt<<1|1);    ma[rt] = max(ma[rt<<1],ma[rt<<1|1]);}int main(){    //freopen("D:\\rush.txt","r",stdin);    int T;    scanf("%d",&T);    while (T--)    {        memset(ma,0,sizeof(ma));        memset(tag,0,sizeof(tag));        dic.clear();        a.clear();        int n;        scanf("%d",&n);        rep1(i,1,n)        {            scanf("%d%d",&aa[i].l,&aa[i].r);            if (aa[i].l>aa[i].r)                swap(aa[i].l,aa[i].r);            if (!dic[aa[i].l])            {                a.push_back(aa[i].l);                dic[aa[i].l] = 1;            }            if (!dic[aa[i].r])            {                a.push_back(aa[i].r);                dic[aa[i].r] = 1;            }        }        sort(a.begin(),a.end());        rep1(i,1,n)        {            int l,r;            l = lower_bound(a.begin(),a.end(),aa[i].l)-a.begin()+1;            r = lower_bound(a.begin(),a.end(),aa[i].r)-a.begin()+1;            up_data(l,r,1,MAXN<<1,1);        }        printf("%d\n",ma[1]);    }    return 0;}