bzoj4378 [POI2015]Logistyka

　　对于询问(a,b)，设X=(数列中小于等于b的数字和),Y=b*(数列中大于b的数字的个数),若X+Y>=a*b，则询问可行，否则不可行，实现的话用两个树状数组维护一下即可

　　代码

　　

#include<cstdio>
#include<algorithm>
#include<set>
#define fi first
#define sc second
#define mp make_pair
#define N 2000100
using namespace std;
long long c1[N],c2[N],e[N],a[N],b[N];
int typ[N],n,i,m,tot,u,v[N];
char ch;
int ef(int x)
{
    int l=1,r=n,m;
    while (l<=r)
    {
        m=(l+r)>>1;
        if (e[m]>x) r=m-1;else l=m+1;
    }
    return r;
}
int lowbit(int x)
{
    return x&-x;
}
void cc(int x,int w,long long ww)
{
    while (x<=n)
    {
        c1[x]+=w;
        c2[x]+=ww;
        x+=lowbit(x);
    }
}
pair<int,long long> sum(int x)
{
    int ans1=0;
    long long ans2=0;
    while (x)
    {
        ans1+=c1[x];
        ans2+=c2[x];
        x-=lowbit(x);
    }
    return mp(ans1,ans2);
}
int main()
{
    scanf("%d%d",&n,&m);
    tot++;e[tot]=0;
    for (i=1;i<=m;i++)
    {
        scanf(" %c %lld%lld",&ch,&a[i],&b[i]);
        if (ch=='U') 
        {
            tot++;e[tot]=b[i];
            typ[i]=0;
        }
        else typ[i]=1;
    }
    sort(e+1,e+1+tot);e[0]=-1;n=0;
    for (i=1;i<=tot;i++)
    if (e[i]!=e[i-1]) e[++n]=e[i];
    cc(1,n,n*e[1]);
    for (i=1;i<=m;i++)
    {
        if (!typ[i])
        {
            u=ef(v[a[i]]);
            cc(u,-1,-v[a[i]]);
            v[a[i]]=b[i];
            u=ef(v[a[i]]);
            cc(u,1,v[a[i]]);
        }
        else
        {
            u=ef(b[i]);
            pair<int,long long> ans2=sum(u);
            long long v1=n-ans2.fi;
            long long v2=ans2.sc;
            if (v1*b[i]+v2>=a[i]*b[i])
            printf("TAK
");
            else
            printf("NIE
");
        }
    }
}