p1245

理解题意就还好吧:刚开始区间内每一个点权值为是s.对于每一个申请,如果能卖就输出yes并卖票,否则输出no.

这不是模拟么?策略都在了.然而c<=30000是会超时的,我们需要一个高级数据结构.

判断能不能卖=询问区间最小值,这个可以用线段树的.而线段树也支持区间修改,这道题就可以开始写了.

输入的 右端点要-1,因为右端点就下车了.

using namespace std;
inline int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') f=-1;ch=getchar();}
    while(isdigit(ch)) {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
using namespace std;
int i,tn,tl,tr;
int minn[400010],reduce[400010];
int c,s,r;
void build(int now,int l,int r)
{
    minn[now]=s;
    if(l==r)
        return;
    int mid=(l+r)/2;
    build(now<<1,l,mid);
    build((now<<1)+1,mid+1,r);
}

void pushdown(int now)
{
    if(reduce[now])
    {
        reduce[now<<1]+=reduce[now];
        reduce[(now<<1)+1]+=reduce[now];
        minn[now<<1]-=reduce[now];
        minn[(now<<1)+1]-=reduce[now];
        reduce[now]=0;
    }
}

void change(int now,int l,int r)
{
    if(tl<=l&&r<=tr)
    {
        minn[now]-=tn;
        reduce[now]+=tn;
        return;
    }
    int mid=(l+r)/2;
    int lc=now<<1;
    int rc=(now<<1)+1;
    pushdown(now);
    if(tl<=mid)
        change(lc,l,mid);
    if(tr>mid)
        change(rc,mid+1,r);
    minn[now]=min(minn[lc],minn[rc]);
}

int ask(int now,int l,int r)
{
    if(tl<=l&&r<=tr)
        return minn[now];
    int mid=(l+r)/2;
    int lc=now<<1;
    int rc=(now<<1)+1;
    pushdown(now);
    int ans=60010;
    if(tl<=mid)
        ans=ask(lc,l,mid);
    if(tr>mid)
        ans=min(ans,ask(rc,mid+1,r));
    return ans;
}

int main()
{
    cin>>c>>s>>r;
    build(1,1,c);
    for(i=1;i<=r;i++)
    {
        cin>>tl>>tr>>tn;
        tr--;
        if(ask(1,1,c)>=tn)
        {
            cout<<"YES"<<endl;
            change(1,1,c);
        }
        else
            cout<<"NO"<<endl;
    }
    return 0;
}