analysis
For a message \ (, Y \ X) , between the two will bovine height \ (- 1 \) .
Consider the difference, modifying the interval into a single site modification
Note to eliminate duplicate messages
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
const int N=10004;
int n,r,h;//I 省略
int c[N];
pair<int,int> a[N];
int main(){scanf("%d%*d%d%d",&n,&h,&r);// 忽略 I
for(int i=1;i<=r;i++){
int x,y;
scanf("%d%d",&x,&y);
if(x>y)x^=y^=x^=y;
a[i].first=x,a[i].second=y;
}
sort(a+1,a+r+1);
for(int i=1;i<=r;i++){if(a[i]==a[i-1])continue;// 去重
c[a[i].first+1]--,c[a[i].second]++;
}
for(int i=1;i<=n;i++){c[i]+=c[i-1];
printf("%d\n",h+c[i]);
}
return 0;
}