BZOJ 3688 折线统计

Posted 2020-08-15 ziliuziliu

dp[i][j][0/1]一下，然后发现可以BIT搞。注意外层for所有点。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 50050
#define mod 100007
using namespace std;
int n,k,f[maxn][12][2],t[maxn<<1][12][2],mx=0;
struct pnt
{
    int x,y;
}p[maxn];
bool cmp(pnt x,pnt y)
{
    return x.x<y.x;
}
int lowbit(int x)
{
    return (x&(-x));
}
void modify(int x,int a,int b,int val)
{
    for (int i=x;i<=mx;i+=lowbit(i))
        t[i][a][b]=(t[i][a][b]+val)%mod;
}
int ask(int x,int a,int b)
{
    int ret=0;
    for (int i=x;i>=1;i-=lowbit(i))
        ret=(ret+t[i][a][b])%mod;
    return ret;
}
int main()
{
    scanf("%d%d",&n,&k);
    for (int i=1;i<=n;i++)
    {
        scanf("%d%d",&p[i].x,&p[i].y);
        mx=max(mx,p[i].y);
    }
    sort(p+1,p+n+1,cmp);
    modify(p[1].y,0,0,1);modify(p[1].y,0,1,1);
    for (int i=2;i<=n;i++)
    {
        for (int j=1;j<=k;j++)
        {
            f[i][j][0]=ask(p[i].y-1,j-1,1)+ask(p[i].y-1,j,0);f[i][j][0]%=mod;
            f[i][j][1]=ask(mx,j,1)-ask(p[i].y,j,1)+ask(mx,j-1,0)-ask(p[i].y,j-1,0);
            f[i][j][1]%=mod;if (f[i][j][1]<0) f[i][j][1]+=mod;
            modify(p[i].y,j,0,f[i][j][0]);modify(p[i].y,j,1,f[i][j][1]);
        }
        modify(p[i].y,0,0,1);modify(p[i].y,0,1,1);
    }
    int ans=0;
    for (int i=1;i<=n;i++) ans=(ans+f[i][k][0]+f[i][k][1])%mod;
    printf("%d\n",ans);
    return 0;
}