设f[i][j]为在坐标(i,j)时的最小点击次数，转移的话从i-1向上飞或者向下掉转移，注意转移完了之后要把管子部分赋值为inf
#include<iostream>
#include<cstdio>
using namespace std;
const int N=10005,inf=1e9;
int n,m,k,p,ans,a[N],b[N],x[N],y[N],f[N][1005];
int main()
{
scanf("%d%d%d",&n,&m,&k);
a[n]=m+1;
for(int i=1;i<=n;i++)
scanf("%d%d",&x[i],&y[i]),a[i]=m+1;
for(int i=1;i<=k;i++)
scanf("%d",&p),scanf("%d%d",&b[p],&a[p]);
for(int i=1;i<=n;i++)
for(int j=0;j<=m;j++)
f[i][j]=inf;
f[0][0]=inf;
for(int i=1;i<=n;i++)
{
for(int j=x[i];j<=m;j++)
f[i][j]=min(f[i][j],min(f[i-1][j-x[i]]+1,f[i][j-x[i]]+1));
for(int j=m-x[i];j<=m;j++)
f[i][m]=min(f[i][m],min(f[i-1][j]+1,f[i][j]+1));
for(int j=b[i]+1;j<=a[i]-1;j++)
if(j+y[i]<=m)
f[i][j]=min(f[i][j],f[i-1][j+y[i]]);
for(int j=1;j<=b[i];j++)
f[i][j]=inf;
for(int j=a[i];j<=m;j++)
f[i][j]=inf;
}
ans=inf;
int cnt=k;
for(int i=n;i>=1;i--)
{
for(int j=b[i]+1;j<=a[i]-1;j++)
ans=min(ans,f[i][j]);
if(ans!=inf)
break;
if(a[i]!=m+1)
cnt--;
}
if(cnt==k)
printf("1
%d
",ans);
else
printf("0
%d
",cnt);
return 0;
}