• 【魔改】树状数组 牛客多校第五场I vcd 几何+阅读理解


    https://www.nowcoder.com/acm/contest/143/I

    vc-dimension 

    题解:分三种情况,组合数学算一下,其中一种要用树状数组维护

    技巧(来自UESTC):1.循环技巧i主j滑

    2.树状数组:一个数列从左到右分别维护某个元素左边比它大num的与右边比他大的num时,从上往下扫, 对每个点的x坐标离散化累加1到X轴上,然后就会发现sum(p[i].x-1)就是左边比它大的,i-1-sum(p[i])就是右边比它大的。 注意y相同的点,需要一起更新(我已开始一个一个更新,根本写不出来)

    orz太屌了

    坑点:我循环写错了+公式里i,j写反了,wa了20发

    比赛时x,y轴都看反了orz

    #define _CRT_SECURE_NO_WARNINGS
    #include<cmath>
    #include<iostream>
    #include<stdio.h>
    #include<algorithm>
    #include<cstring>
    #include<stack>
    #include<vector>
    #include<string.h>
    #include<queue>
    #include<string>
    #include<set>
    using namespace std;
    #define rep(i,t,n)  for(int i =(t);i<=(n);++i)
    #define per(i,n,t)  for(int i =(n);i>=(t);--i)
    #define mmm(a,b) memset(a,b,sizeof(a))
    #define eps 1e-6
    #define pb push_back
    #define lowbit(x) ((x)&(-(x)))
    const int maxn = 3e5 + 5;
    const int inf = 1e7 + 5;//0x7fffffff;   //无限大
    const int MOD = 1013;
    typedef long long ll;
    const int mod = 998244353;
    int power(int a, int b) {
        int c = 1; a %= mod;
        while (b) {
            if (b & 1) c = 1ll * c*a%mod;
            a = 1ll * a*a%mod; b >>= 1;
        }
        return c;
    }
    
    
    struct node {
        int x, y;
        void sc() { scanf("%d%d", &x, &y); }
    }p[maxn];
    
    int n;
    bool cmp(node a, node b) {
        return a.x > b.x;
    }
    
    
    bool cmpy(node a, node b) {
        return a.y < b.y;
    }
    
    int f[maxn];
    int idx;
    void add(int x, int y) { for (; x <= idx; x += lowbit(x)) f[x] += y; }
    int sum(int x) { int ans = 0; for (; x; x -= lowbit(x)) ans += f[x]; return ans; }
    
    int main() {
        int div2 = power(2, mod - 2);
        cin >> n;
        rep(i, 1, n) {
            p[i].sc();
        }
        int ans = n;
         ans +=  1ll*n * (n - 1)%mod*div2 % mod;
        if (ans >= mod)ans -= mod; else if (ans < 0)ans += mod;
        sort(p + 1, p + 1 + n, cmpy);
         idx = 0;
        for (int i = 1, j; i <= n; i = j + 1) {
            for (j = i ; j < n&&p[i].y == p[j+1].y; j++);
            ++idx;
            rep(k, i, j)p[k].y = idx;
            ans -=1ll* (j - i)*(j - i + 1)%mod*div2%mod;
            if (ans >= mod)ans -= mod; else if (ans < 0)ans += mod;
        }
        mmm(f, 0);
        sort(p + 1, p + 1 + n, cmp);
        for (int i = 1, j; i <= n; i = j + 1) {
            for (j = i ; j < n&&p[i].x == p[j+1].x; j++);
            rep(k, i, j) {
                int t1 = sum(p[k].y - 1); int t2 = i - 1 - sum(p[k].y);
                ans += 1ll * t1*t2%mod;
                if (ans >= mod)ans -= mod; else if (ans < 0)ans += mod;
            }
            rep(k, i, j)add(p[k].y, 1);
        }
        cout << ans << endl;
        cin >> n;
        return 0;
    }
    
    /*
    
    */

    一开始无脑分别前后维护

    #define _CRT_SECURE_NO_WARNINGS
    #include<cmath>
    #include<iostream>
    #include<stdio.h>
    #include<algorithm>
    #include<cstring>
    #include<stack>
    #include<vector>
    #include<string.h>
    #include<queue>
    #include<string>
    #include<set>
    using namespace std;
    #define rep(i,t,n)  for(int i =(t);i<=(n);++i)
    #define per(i,n,t)  for(int i =(n);i>=(t);--i)
    #define mmm(a,b) memset(a,b,sizeof(a))
    #define eps 1e-6
    #define pb push_back
    const int maxn = 1e5 + 5;
    const int inf = 1e7 + 5;//0x7fffffff;   //无限大
    const int MOD = 1013;
    typedef long long ll;
    const long long mod = 998244353;
    int power(int a,int b){
        int c=1; a%=mod;
        while (b) {
            if (b&1) c=1ll*c*a%mod;
            a=1ll*a*a%mod; b>>=1;
        }
        return c;
    }
    
    ll div2 = power(2, mod - 2);
    struct node {
        int x, y;
        int id;
        void sc() { scanf("%d%d", &x, &y); }
    }P[maxn];
    
    ll n;
    bool cmp(node a, node b) {
        if (a.x != b.x)
        return a.x < b.x;
        else return a.y < b.y;
    }
    bool cmpx(node a, node b) {
        if (a.x != b.x)
            return a.x > b.x;
        else return a.y < b.y;
    }
    
    bool cmpy(node a, node b) {
        return a.y < b.y;
    }
    int cnt[maxn];
    ll l[maxn], r[maxn];
    int d[maxn];
    int lowbit(int x) { return x & (-x); }
    int fr[maxn], bk[maxn];
    void add(int x, int v) {//a[x]+=v;
        while (x <= maxn) {
            d[x] += v;
            x += lowbit(x);
        }
    
    }
    int query(int x) {
        int res = 0;
        while (x) {
            res += d[x];
            x -= lowbit(x);
        }
        return res;
    }
    
    int main() {
        cin >> n;
        rep(i, 1, n) {
            P[i].sc();
            P[i].id = i;
        }
        sort(P + 1, P + n + 1, cmp);
    
        int idx = 0;
        int now = 0;
        rep(i, 1, n) {
            if (P[i].x != now) {
                cnt[++idx]++;
                now = P[i].x;
            }
            else {
                cnt[idx]++;
            }
        }
        ll ans = n;
        ll tmp = n * (n - 1) %mod*div2;
        rep(i, 1, idx)if (cnt[i] > 1) { tmp = (tmp - cnt[i] * (cnt[i] - 1) % mod *div2) % mod; }
        ans = (ans + tmp) % mod;
        //维护每个点左右比它高的点
        sort(P + 1, P + n + 1, cmpy);
        idx = 0;
        now = 0;
        rep(i, 1, n) {
            if (P[i].y != now) {
                P[i].y = ++idx;
                now = P[i].y;
            }
            else {
                P[i].y = idx;
            }
        }
        sort(P + 1, P + n + 1, cmp);
         now = 0;
        rep(i, 1, n)
        {
            if (P[i].x != P[i - 1].x)fr[i] =now= i - 1;
            else fr[i] = now;
        }
        now = 0;
        per(i, n, 1)
        {
            if (P[i].x != P[i + 1].x)bk[i] = now = i + 1;
            else bk[i] = now;
        }
        rep(i, 1, n) {
            if(P[i].x!=P[i-1].x)l[i] =ll( i-1-query(P[i].y));
            else {
                if (P[i].y >= P[fr[i]].y || P[i].y >= P[bk[i]].y)l[i] = 0;
                else l[i] = l[i - 1];
            }
            add(P[i].y, 1);    
        }
        mmm(d, 0);
        //sort(P + 1, P + n + 1, cmpx);
        int first = 1;
        per(i, n, 1) {
            if (P[i].x != P[i + 1].x)r[i] = ll(n-i - query(P[i].y)),first=1;
            else {
                if (P[i].y >= P[fr[i]].y || P[i].y >= P[bk[i]].y)r[i] = 0;
                else {
                    if (first) 
                    { 
                        first = 0; 
                        r[i] = ll(n - i -1- query(P[i].y));
                    }
                    else r[i] = r[i + 1]; 
                }
            }
            add(P[i].y, 1);
        }
        tmp = 0;
        rep(i, 1, n) {
            tmp = (tmp + l[i] * r[i] % mod) % mod;
        }
        ans = (ans + tmp) % mod;
            cout << ans<<endl;
    
    
            //sort(P + 1, P + n + 1, cmp);
            rep(i, 1, n) {
                //cout << l[i] << ' ' << r[i] << endl;
            }
        cin >> n;
    }
    
    /*
    
    */
    成功的路并不拥挤,因为大部分人都在颓(笑)
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  • 原文地址:https://www.cnblogs.com/SuuT/p/9410287.html
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