F. Cutting Game

合集 - 模拟(54)

1.G. Vlad and Trouble at MIT2024-02-202.P2899 [USACO08JAN] Cell Phone Network G2024-02-203.Watering an Array2024-02-234.P2055 [ZJOI2009] 假期的宿舍2024-03-085.T432579 大西洋船王2024-03-086.T432580 星云 hard ver.2024-03-087.P7072 [CSP-J2020] 直播获奖2024-03-098.E - Insert or Erase2024-03-109.D - String Bags2024-03-1010.P2671 [NOIP2015 普及组] 求和2024-03-1011.P1884 [USACO12FEB] Overplanting S2024-03-1012.F. Microcycle2024-03-1213.E. Increasing Subsequences__22024-03-1314.G. Path Prefixes2024-03-1415.F. Rudolf and Imbalance2024-03-1416.P1355 神秘大三角2024-03-2017.[ARC172A] Chocolate2024-03-2518.F. 0, 1, 2, Tree!2024-03-2919.E. Nearly Shortest Repeating Substring2024-03-2920.P1354 房间最短路问题2024-03-2921.P3258 [JLOI2014] 松鼠的新家2024-04-0222.P3052 [USACO12MAR] Cows in a Skyscraper G2024-04-0523.P1337 [JSOI2004] 平衡点 / 吊打XXX2024-04-0724.P3745 [六省联考 2017] 期末考试2024-04-0825.D. Inaccurate Subsequence Search2024-04-0926.E. Long Inversions2024-04-0927.C. Deep Down Below2024-04-0928.C. Rings2024-04-0929.F. Unfair Game2024-04-1030.C. Inversion Graph2024-04-1031.D. Cross Coloring2024-04-1232.D2. Set To Max (Hard Version)2024-04-1233.C. Array Game2024-04-1234.F. Alex's whims2024-04-1235.Queue Sort2024-04-1236.P10252 线性变换2024-04-1637.P10288 [GESP样题 八级] 区间2024-04-1638.P2161 [SHOI2009] 会场预约2024-04-1739.L2-001 紧急救援2024-04-1840.L2-002 链表去重2024-04-1841.L2-013 红色警报2024-04-1842.L2-014 列车调度2024-04-1943.L2-022 重排链表2024-04-1944.L2-023 图着色问题2024-04-1945.L2-039 清点代码库2024-04-1946.L2-034 口罩发放2024-04-1947.L3-037 夺宝大赛2024-04-2348.仓库规划2024-04-2349.因子化简2024-04-2350.C. Torn Lucky Ticket2024-04-2751.俄罗斯方块2024-05-14

52.F. Cutting Game2024-05-21

53.D. Prefix Permutation Sums2024-06-1754.P3304 [SDOI2013] 直径2024-07-15

收起

原题链接

题解

我一开始想着二位前缀和+模拟，但是看到ab发现空间不够
观察到nm的数据范围限制
也就是点和查询的数量限制，考虑用优先队列把各个方向的点存起来以表示各个点的使用顺序，再用map嵌套表示各个点的使用情况

空间复杂度 $O(n)$ 时间复杂度 $O(T(mlogn+nlogn))$

code

#include<bits/stdc++.h>
using namespace std;
struct nodel
{
    int x,y;
    bool operator<(const nodel &b)const {return b.y<y;}
};
struct noder
{
    int x,y;
    bool operator<(const noder &b)const {return b.y>y;}
};
struct nodeu
{
    int x,y;
    bool operator<(const nodeu &b)const {return b.x<x;}
};
struct noded
{
    int x,y;
    bool operator<(const noded &b)const {return b.x>x;}
};
map<int,map<int,int > >vis;
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int t;
    cin>>t;
    while(t--)
    {
        int a,b,n,m;
        cin>>a>>b>>n>>m;
        priority_queue<nodel> l;
        priority_queue<noder> r;
        priority_queue<nodeu> u;
        priority_queue<noded> d;

        while(n--)
        {
            int x,y;
            cin>>x>>y;
            vis[x][y]=1;
            l.push({x,y});
            r.push({x,y});
            u.push({x,y});
            d.push({x,y});
        }

        int x1=1,y1=1,x2=a,y2=b;

        int flag=1;
        int ansa=0,ansb=0;
        while(m--)
        {
            char op;
            int k;
            cin>>op>>k;


            int cnt=0;
            if(op=='U')
            {
                x1+=k;
                while(u.size())
                {
                    int x=u.top().x,y=u.top().y;
                    if(x<x1)
                    {
                        if(vis[x][y]) cnt++;
                        vis[x][y]=0;
                        u.pop();
                    }
                    else break;
                }
            }
            else if(op=='D')
            {
                x2-=k;
                while(d.size())
                {
                    int x=d.top().x,y=d.top().y;
                    if(x>x2)
                    {
                        if(vis[x][y]) cnt++;
                        vis[x][y]=0;
                        d.pop();
                    }
                    else break;
                }
            }
            else if(op=='R')
            {
                y2-=k;
                while(r.size())
                {
                    int x=r.top().x,y=r.top().y;
                    if(y>y2)
                    {
                        if(vis[x][y]) cnt++;
                        vis[x][y]=0;
                        r.pop();
                    }
                    else break;
                }
            }
            else
            {
                y1+=k;
                while(l.size())
                {
                    int x=l.top().x,y=l.top().y;
                    if(y<y1)
                    {
                        if(vis[x][y]) cnt++;
                        vis[x][y]=0;
                        l.pop();
                    }
                    else break;
                }
            }

            if(flag) ansa+=cnt;
            else ansb+=cnt;
            flag^=1;
        }
        cout<<ansa<<" "<<ansb<<"\n";
        vis.clear();
    }
    return 0;
}