http://acm.hdu.edu.cn/showproblem.php?pid=5124
求x軸上被線段覆蓋最多次的點的線段覆蓋次數
輸入的左右端點範圍有10^9,離散化一下即可
有點像cf上某題的變種
[xi,yi]分爲兩個端點xi和(yi)+1,在xi時該點會新加入一條線段,同樣的,在(yi)+1時該點會減少一條線段,對每條線段,使xi++,yi--,最後將所有出現的端點排序後求最大前綴和即可。
<span style="font-size:14px;">#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <sstream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define clr0(x) memset(x,0,sizeof(x))
#define clr1(x) memset(x,-1,sizeof(x))
#define eps 1e-9
const double pi = acos(-1.0);
typedef long long LL;
const int inf = 1000000000;
const int maxn = 2e5 + 5;
map<int,int > hash;
#define x first
#define y second
typedef pair<int,int> p2;
p2 s[maxn];
int a[maxn];
int work()
{
int n,x0,y0,cnt_a = 0;
RD(n);
for(int i = 0;i < n;++i){
RD2(s[i].x,s[i].y);
s[i].y++;
a[cnt_a++] = s[i].x;
a[cnt_a++] = s[i].y;
}
int ans = 1;
sort(a,a+cnt_a);
int m = 1;
hash[a[0]] = m++;
for(int i = 1;i < cnt_a;++i){
if(a[i] != a[i - 1])
hash[a[i]] = m++;
}
clr0(a);
for(int i = 0;i < n;++i){
a[hash[s[i].x]]++,a[hash[s[i].y]]--;
}
int res = 0;
for(int i = 0;i < cnt_a;++i){
res+= a[i];
ans = max(ans,res);
}
return ans;
}
int main()
{
int _;
RD(_);
while(_--){
printf("%d\n",work());
}
return 0;
}
</span>