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  • BZOJ 2527 Meteors

    整体二分:二分操作。

    首先,memset很慢。

    其次,最好开long long。

    然后,最后再下一场inf流星雨保证能够出解。

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #define maxn 300500
    #define inf 1000000010
    using namespace std;
    long long n,m,k,o[maxn],p[maxn],id[maxn],nxt[maxn],g[maxn],val[maxn],ans[maxn];
    long long cur[maxn],tmp[maxn];
    long long x,y,z,tol[maxn],tor[maxn];
    struct meteor
    {
        long long l,r,a;
    }q[maxn];
    void addcon(long long con,long long now)
    {
        nxt[now]=g[con];
        g[con]=now;
    }
    void addmet(long long left,long long right,long long a,long long now)
    {
        q[now].l=left;
        q[now].r=right;
        q[now].a=a;
    }
    long long lowbit(long long x)
    {
        return ((-x)&x);
    }
    void modify(long long now,long long x)
    {
        for (long long i=now;i<=m;i+=lowbit(i))
            val[i]+=x;
    }
    void modifys(long long now,long long x)
    {
        if (q[now].l<=q[now].r) 
        {
            modify(q[now].l,x);
            if (q[now].r+1<=m) modify(q[now].r+1,-x);
        }
        else
        {
            modify(q[now].l,x);
            modify(1,x);
            modify(q[now].r+1,-x);
        }
    }
    long long query(long long x)
    {
        long long ret=0;
        for (long long i=x;i>=1;i-=lowbit(i))
            ret+=val[i];
        return ret;
    }
    void half_solve(long long head,long long tail,long long left,long long right)
    {
        if (head>tail) return;
        long long lsum=0,rsum=0;
        if (left==right)
        {
            for (long long i=head;i<=tail;i++)
                ans[id[i]]=left;
            return;
        }
        long long mid=(left+right)>>1;
        for (long long i=left;i<=mid;i++)
            modifys(i,q[i].a);
        for (long long i=head;i<=tail;i++)
        {
            tmp[id[i]]=0;
            for (long long j=g[id[i]];j;j=nxt[j])
            {
                tmp[id[i]]+=query(j);
                if (cur[id[i]]+tmp[id[i]]>p[id[i]]) break;
            }
            if (cur[id[i]]+tmp[id[i]]>=p[id[i]]) tol[++lsum]=id[i];
            else {cur[id[i]]+=tmp[id[i]];tor[++rsum]=id[i];}
        }
        for (long long i=left;i<=mid;i++)
            modifys(i,-q[i].a);
        for (long long i=1;i<=lsum;i++) id[head-1+i]=tol[i];
        for (long long i=1;i<=rsum;i++) id[head-1+lsum+i]=tor[i];
        half_solve(head,head+lsum-1,left,mid);
        half_solve(head+lsum,tail,mid+1,right);
    }
    int main()
    {
        scanf("%lld%lld",&n,&m);
        for (long long i=1;i<=m;i++)
        {
            scanf("%lld",&o[i]);
            addcon(o[i],i);
        }
        for (long long i=1;i<=n;i++)
        {
            scanf("%lld",&p[i]);
            id[i]=i;
        }
        scanf("%lld",&k);
        for (long long i=1;i<=k;i++)
        {
            scanf("%lld%lld%lld",&x,&y,&z);
            addmet(x,y,z,i);
        }
        k++;addmet(1,n,inf,k);
        half_solve(1,n,1,k);
        for (long long i=1;i<=n;i++)
        {
            if (ans[i]==k) printf("NIE
    ");
            else printf("%lld
    ",ans[i]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/ziliuziliu/p/5514992.html
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