• 模拟赛小结:2019-2020 ICPC, Asia Jakarta Regional Contest


    比赛链接:传送门

    离金最近的一次?,lh大佬carry场。

    Problem A. Copying Homework 00:17(+) Solved by Dancepted

    签到,读题有点慢了。而且配置vscode花了点时间。

    #include <bits/stdc++.h>
     
    using namespace std;
     
    int a[100005];
    int main() {
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }
        for (int i = 1; i <= n; i++) {
            printf("%d%c", n + 1 - a[i], " 
    "[i==n]);
        }
        return 0;
    }
    View Code

    Problem C. Even Path 00:44(+) Solved by Dancepted

    每个格子通过even path能到达的区域是一个矩形,预处理这个矩形的边界就ok。签到。

    “NO”写成了“No”,WA1不算罚时开心死了。

    #include <bits/stdc++.h>
    #define N 100005
     
    using namespace std;
     
    int R[N], C[N];
    int l[N], r[N], u[N], d[N];
     
    int main() {
        int n, q;
        cin >> n >> q;
        for (int i = 1; i <= n; i++)
            scanf("%d", &R[i]);
        for (int i = 1; i <= n; i++)
            scanf("%d", &C[i]);
        l[1] = u[1] = 1;
        for (int i = 2; i <= n; i++) {
            if (R[i]%2 == R[i-1]%2)
                u[i] = u[i - 1];
            else
                u[i] = i;
            if (C[i]%2 == C[i-1]%2)
                l[i] = l[i - 1];
            else
                l[i] = i;
        }
        r[n] = d[n] = n;
        for (int i = n - 1; i >= 1; i--) {
            if (R[i]%2 == R[i+1]%2)
                d[i] = d[i + 1];
            else
                d[i] = i;
            if (C[i]%2 == C[i+1]%2)
                r[i] = r[i + 1];
            else
                r[i] = i;
        }
        while (q--) {
            int ra, ca, rb, cb;
            scanf("%d%d%d%d", &ra, &ca, &rb, &cb);
            if (u[ra] <= rb && rb <= d[ra] && l[ca] <= cb && cb <= r[ca])
                puts("YES");
            else
                puts("NO");
        }
        return 0;
    }
    View Code

    Problem H. Twin Buildings 01:03(-1) Solved by xk

    xk说是sb题。。。qwq。

    #include<bits/stdc++.h>
    using namespace std;
    #define forn(i, n) for(int i = 0; i < (n); i++)
    #define forab(i, a, b) for(int i = (a); i <= (b); i++)
    #define forba(i, b, a) for(int i = (b); i >= (a); i--)
    #define mst(array, Num, Kind, Count) memset(array, Num, sizeof(Kind) * (Count))
    #define mp(x, y) make_pair(x, y)
    #define fi first
    #define se second
    typedef long long ll;
    typedef double db;
    typedef pair<int, int> pii;
     
    const int maxn = 1e5 + 5;
     
    pii a[maxn];
    int mxh[maxn];
     
    int main()
    {
        int n;
        ios::sync_with_stdio(0), cin.tie(0);
        cin >> n;
        ll ans = 0;
        forn(i, n)
        {
            cin >> a[i].fi >> a[i].se;
            if(a[i].fi > a[i].se)
                swap(a[i].fi, a[i].se);
            ans = max(ans, (ll)a[i].fi * a[i].se);
        }
        sort(a, a + n);
        forba(i, n - 1, 1)
        {
            mxh[i - 1] = max(mxh[i], a[i].se);
            ans = max(ans, (ll)a[i - 1].fi * min(a[i - 1].se, mxh[i - 1]) * 2);
        }
        cout << ans / 2 << (ans % 2 ? ".5" : ".0") << endl;
    }
    View Code

    Problem K. Addition Robot 01:28(+) Solved by lh

    lh一拍脑袋说这个tm好像可以区间合并,那不是线段树随便搞?于是K就被秒了。

    具体地:

    若左子树对应区间的A和B是:

    $egin{bmatrix}A_{l}&B_{l}end{bmatrix} = egin{bmatrix}A&Bend{bmatrix} * egin{bmatrix}f_{AtoAl} & f_{AtoBl}\ f_{BtoAl} & f_{BtoBl}end{bmatrix}$

    而右子树对应区间的A和B是:

    $egin{bmatrix}A_{r}&B_{r}end{bmatrix} = egin{bmatrix}A&Bend{bmatrix} * egin{bmatrix}f_{AtoAr} & f_{AtoBr}\ f_{BtoAr} & f_{BtoBr}end{bmatrix}$

    则合并后的区间对应的A和B是:

    $egin{bmatrix}A_{o}&B_{o}end{bmatrix} = egin{bmatrix}A&Bend{bmatrix} * egin{bmatrix}f_{AtoAl} & f_{AtoBl}\ f_{BtoAl} & f_{BtoBl}end{bmatrix} * egin{bmatrix}f_{AtoAr} & f_{AtoBr}\ f_{BtoAr} & f_{BtoBr}end{bmatrix}$

    线段树每个节点维护4个系数。对于交换一个区间内的A和B,就是把系数AtoA和BtoA交换,AtoB和BtoB交换。合并的时候两个子区间的系数矩阵乘一下就行。

    #include<bits/stdc++.h>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define LLINF 0x3f3f3f3f3f3f3f3f
    #define lowbit(x) ((-x) & x)
    #define ffor(i, d, u) for (int i = (d); i <= (u); ++i)
    #define _ffor(i, u, d) for (int i = (u); i >= (d); --i)
    #define mst(array, Num, Kind, Count) memset(array, Num, sizeof(Kind) * (Count))
    #define mp(x, y) make_pair(x, y)
    #define fi first
    #define se second
    #define N 100005
    #define M 1000005
    typedef long long ll;
    typedef double db;
    typedef pair<ll, ll> pll;
    typedef pair<int, int> pii;
    typedef pair<db, db> pdd;
    const db PI = acos(-1);
    const ll MO = 1e9 + 7;
    const ll Inv2 = (MO + 1) / 2;
    const bool debug = true;
    template <typename T>
    inline void read(T &x)
    {
        x=0;char c;T t=1;while(((c=getchar())<'0'||c>'9')&&c!='-');
        if(c=='-'){t=-1;c=getchar();}do(x*=10)+=(c-'0');while((c=getchar())>='0'&&c<='9');x*=t;
    }
    template <typename T, typename... Args>
    inline void read(T &x, Args &... args)
    {
        read(x), read(args...);
    }
    template <typename T>
    inline void write(T x)
    {
        int len=0;char c[21];if(x<0)putchar('-'),x*=(-1);
        do{++len;c[len]=(x%10)+'0';}while(x/=10);_ffor(i,len,1)putchar(c[i]);
    }
    int n, q;
    char s[N];
    struct node
    {
        ll a, b, x, y;
        bool lazy;
        node operator * (const node &other)
        {
            node ans;
            ans.lazy = false;
            ans.a = (a * other.a % MO + x * other.b % MO) % MO;
            ans.b = (b * other.a % MO + y * other.b % MO) % MO;
            ans.x = (a * other.x % MO + x * other.y % MO) % MO;
            ans.y = (b * other.x % MO + y * other.y % MO) % MO;
            return ans;
        }
    } t[N << 2];
    inline void pushup(int o)
    {
        int lo = o << 1, ro = o << 1 | 1;
        t[o] = t[lo] * t[ro];
    }
    void build(int o = 1, int l = 1, int r = n)
    {
        t[o].lazy = false;
        if (l == r)
        {
            if (s[l] == 'A')
                t[o].a = 1, t[o].b = 1, t[o].x = 0, t[o].y = 1;
            else
                t[o].a = 1, t[o].b = 0, t[o].x = 1, t[o].y = 1;
            return;
        }
        int mid = (l + r) >> 1;
        build(o << 1, l, mid), build(o << 1 | 1, mid + 1, r);
        pushup(o);
    }
    inline void modify(int o)
    {
        t[o].lazy = !t[o].lazy;
        swap(t[o].a, t[o].b), swap(t[o].x, t[o].y);
        swap(t[o].a, t[o].x), swap(t[o].b, t[o].y);
    }
    inline void pushdown(int o)
    {
        if (t[o].lazy == false)
            return;
        int lo = o << 1, ro = o << 1 | 1;
        modify(lo), modify(ro), t[o].lazy = false;
    }
    void change(int cl, int cr, int o = 1, int l = 1, int r = n)
    {
        if (cl == l && cr == r)
        {
            modify(o);
            return;
        }
        pushdown(o);
        int mid = (l + r) >> 1;
        if (mid >= cr)
            change(cl, cr, o << 1, l, mid);
        else if (mid < cl)
            change(cl, cr, o << 1 | 1, mid + 1, r);
        else
            change(cl, mid, o << 1, l, mid), change(mid + 1, cr, o << 1 | 1, mid + 1, r);
        pushup(o);
    }
    node query(int ql, int qr, int o = 1, int l = 1, int r = n)
    {
        if (ql == l && qr == r)
            return t[o];
        pushdown(o);
        int mid = (l + r) >> 1;
        if (mid >= qr)
            return query(ql, qr, o << 1, l, mid);
        else if (mid < ql)
            return query(ql, qr, o << 1 | 1, mid + 1, r);
        else
            return query(ql, mid, o << 1, l, mid) * query(mid + 1, qr, o << 1 | 1, mid + 1, r);
    }
    inline int ac()
    {
        read(n, q);
        scanf("%s", s + 1);
        build();
        while (q--)
        {
            int op, l, r, a, b;
            read(op, l, r);
            if (op == 2)
            {
                read(a, b);
                node ans = query(l, r);
                write((a * ans.a % MO + b * ans.b % MO) % MO), putchar(' '), write((a * ans.x % MO + b * ans.y % MO) % MO), putchar('
    ');
            }
            else
                change(l, r);
        }
        return 0;
    }
    int main()
    {
        ac();
        return 0;
    }
    View Code

    Problem G. Performance Review  02:44(+) Solved by lh

    据说是个线段树之类的题,lh分分钟就秒了,我和xk站在一边人都看傻了。

    #include<bits/stdc++.h>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define lowbit(x) ((-x) & x)
    #define ffor(i, d, u) for (int i = (d); i <= (u); ++i)
    #define _ffor(i, u, d) for (int i = (u); i >= (d); --i)
    #define mst(array, Num, Kind, Count) memset(array, Num, sizeof(Kind) * (Count))
    #define mp(x, y) make_pair(x, y)
    #define fi first
    #define se second
    #define N 100005
    #define M 3000005
    typedef long long ll;
    typedef double db;
    typedef pair<ll, ll> pll;
    typedef pair<int, int> pii;
    typedef pair<db, db> pdd;
    const db PI = acos(-1);
    const ll MO = 1e9 + 7;
    const ll Inv2 = (MO + 1) / 2;
    const bool debug = true;
    template <typename T>
    inline void read(T &x)
    {
        x=0;char c;T t=1;while(((c=getchar())<'0'||c>'9')&&c!='-');
        if(c=='-'){t=-1;c=getchar();}do(x*=10)+=(c-'0');while((c=getchar())>='0'&&c<='9');x*=t;
    }
    template <typename T, typename... Args>
    inline void read(T &x, Args &... args)
    {
        read(x), read(args...);
    }
    template <typename T>
    inline void write(T x)
    {
        int len=0;char c[21];if(x<0)putchar('-'),x*=(-1);
        do{++len;c[len]=(x%10)+'0';}while(x/=10);_ffor(i,len,1)putchar(c[i]);
    }
    int n, m, q, a = 0;
    vector<int> b[N];
    int minx[N << 2], lazy[N << 2];
    void build(int o = 1, int l = 1, int r = m)
    {
        minx[o] = a, lazy[o] = 0;
        if (l == r)
            return;
        int mid = (l + r) >> 1;
        build(o << 1, l, mid), build(o << 1 | 1, mid + 1, r);
    }
    inline void pushdown(int o)
    {
        if (lazy[o] == 0)
            return;
        int lo = o << 1, ro = o << 1 | 1;
        minx[lo] += lazy[o], minx[ro] += lazy[o];
        lazy[lo] += lazy[o], lazy[ro] += lazy[o];
        lazy[o] = 0;
    }
    void change(int ql, int qr, int x, int o = 1, int l = 1, int r = m)
    {
        if (l == ql && r == qr)
        {
            lazy[o] += x, minx[o] += x;
            return;
        }
        pushdown(o);
        int mid = (l + r) >> 1;
        if (mid >= qr)
            change(ql, qr, x, o << 1, l, mid);
        else if (mid < ql)
            change(ql, qr, x, o << 1 | 1, mid + 1, r);
        else
            change(ql, mid, x, o << 1, l, mid), change(mid + 1, qr, x, o << 1 | 1, mid + 1, r);
        minx[o] = min(minx[o << 1], minx[o << 1 | 1]);
    }
    inline int ac()
    {
        int r, x, y, z;
        read(n, m, q, x);
        ffor(i, 2, n) read(y), a += (y < x ? 1 : 0);
        build(), a = 0;
        ffor(i, 1, m)
        {
            read(r);
            if (a)
                change(i, m, a);
            a = 0;
            ffor(j, 1, r)
            {
                read(y), b[i].emplace_back(y);
                a += (y < x ? 1 : 0);
            }
            change(i, m, -r);
        }
        int mu = x;
        while (q--)
        {
            read(x, y, z);
            if ((b[x][y - 1] > mu) == (z > mu) || x == m)
            {
                b[x][y - 1] = z;
                puts(minx[1] >= 0 ? "1" : "0");
                continue;
            }
            if (z > mu)
                change(x + 1, m, -1);
            else
                change(x + 1, m, 1);
            b[x][y - 1] = z;
            puts(minx[1] >= 0 ? "1" : "0");
        }
        return 0;
    }
    int main()
    {
        ac();
        return 0;
    }
    View Code

    后面我和xk去开了E,我想了个贪心,但是有些细节没理清楚,而xk怎么看怎么像差分约束,对着板子敲了一下,但是后面发现不太会改板子的样子。后面还尝试写F和L。

    F我看了一眼割出来的点必须是树的重心,但是不知道怎么判子树同构,瞎猜了一个做法,但是bug有点多没来得及调完。(赛后调完交上去WA17,应该是个假做法,明早补)

    L大概是个图论,但是xk最后没来得及写。


    总结:

    这场其实有点不太认真,因为各种原因过了10分钟左右才开始看题。中间我还接到了移动的电话骗我去改套餐(结果是兼职的同学找人代打电话,细节没交代好,害我白跑一趟),还顺路买了个晚饭。。。

    我后面上厕所也上了挺久的,浪费了不少时间。

    最后手上还有三道疑似可做的题,时间多一点说不定能再过一题,这周六要认真点打了QAQ。

    还有个人感觉xk看E的时候的:“这肯定是个差分约束,但是我不会差分约束”,这种做法不太好,我感觉这样很容易看成假算法。我有一场CF就是死脑筋,那场的D我就说:“这肯定是个马拉车,但是我不会马拉车,我去看一下我的板子”,但是实际上D是个思维题,我还因此掉了不少分。

    不过xk和我切水题又快起来了的样子qwq。然后lh一眼一道线段树,分分钟过掉真的nb没话说。

    赛后想想,F题的判树的同构的算法是猜出来的,而xk的L应该是证明过的比较严谨,下次碰到的时候应该优先让这种题先敲。

    其他的。。。和某位肾结石的好基友讨论了一下,才发现我们俩从初中开始就熬夜成瘾了,身体真的吃不消,以后还是早睡早起吧QWQ。

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  • 原文地址:https://www.cnblogs.com/Lubixiaosi-Zhaocao/p/11774026.html
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