Codeforces 283E Cow Tennis Tournament 线段树 （看题解)

Cow Tennis Tournament

感觉这题的难点在于想到求违反条件的三元组。。

为什么在自己想的时候没有想到求反面呢！！！！

违反的三元组肯定存在一个人能打败其他两个人， 扫描的过程中用线段树维护一下就好了。

反思： 计数问题： 正难则反 正难则反 正难则反 ！！！！

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, k, s[N];
int cntL[N], cntR[N];
vector<int> in[N], ot[N];

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1

struct setmentTree {
    int a[N << 2][2];
    int flip[N << 2];

    inline void pull(int rt) {
        a[rt][0] = a[rt << 1][0] + a[rt << 1 | 1][0];
        a[rt][1] = a[rt << 1][1] + a[rt << 1 | 1][1];
    }
    inline void push(int rt) {
        if(flip[rt]) {
            swap(a[rt << 1][0], a[rt << 1][1]);
            swap(a[rt << 1 | 1][0], a[rt << 1 | 1][1]);
            flip[rt << 1] ^= 1;
            flip[rt << 1 | 1] ^= 1;
            flip[rt] = 0;
        }
    }
    void build(int l, int r, int rt) {
        flip[rt] = 0;
        if(l == r) {
            a[rt][0] = 1;
            a[rt][1] = 0;
            return;
        }
        int mid = l + r >> 1;
        build(lson); build(rson);
        pull(rt);
    }
    void update(int L, int R, int l, int r, int rt) {
        if(R < l || r < L || R < L) return;
        if(L <= l && r <= R) {
            swap(a[rt][0], a[rt][1]);
            flip[rt] ^= 1;
            return;
        }
        push(rt);
        int mid = l + r >> 1;
        update(L, R, lson);
        update(L, R, rson);
        pull(rt);
    }
    PII query(int L, int R, int l, int r, int rt) {
        if(R < l || r < L || R < L) return mk(0, 0);
        if(L <= l && r <= R) return mk(a[rt][0], a[rt][1]);
        push(rt);
        int mid = l + r >> 1;
        PII ret, tmp;
        tmp = query(L, R, lson);
        ret.fi += tmp.fi; ret.se += tmp.se;
        tmp = query(L, R, rson);
        ret.fi += tmp.fi; ret.se += tmp.se;
        return ret;
    }
} Tree;

struct Line {
    int l, r;
} seg[N];

int main() {
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++) scanf("%d", &s[i]);
    sort(s + 1, s + 1 + n);
    for(int i = 1; i <= k; i++) {
        int a, b; scanf("%d%d", &a, &b);
        a = lower_bound(s + 1, s + 1 + n, a) - s;
        b = upper_bound(s + 1, s + 1 + n, b) - s - 1;
        seg[i].l = a; seg[i].r = b;
        if(a <= b) {
            in[a].push_back(i);
            ot[b].push_back(i);
        }
    }
    Tree.build(1, n, 1);
    for(int i = 1; i <= n; i++) {
        for(auto &id : in[i]) Tree.update(seg[id].l, seg[id].r, 1, n, 1);
        cntL[i] = Tree.query(1, i - 1, 1, n, 1).fi;
        for(auto &id : ot[i]) Tree.update(seg[id].l, seg[id].r, 1, n, 1);
    }
    Tree.build(1, n, 1);
    for(int i = n; i >= 1; i--) {
        for(auto &id : ot[i]) Tree.update(seg[id].l, seg[id].r, 1, n, 1);
        cntR[i] = Tree.query(i + 1, n, 1, n, 1).se;
        for(auto &id : in[i]) Tree.update(seg[id].l, seg[id].r, 1, n, 1);
    }
    LL ans = 1LL * n * (n - 1) * (n - 2) / 6;
    for(int i = 1; i <= n; i++) {
        ans -= 1LL * cntL[i] * (cntL[i] - 1) / 2;
        ans -= 1LL * cntR[i] * (cntR[i] - 1) / 2;
        ans -= 1LL * cntL[i] * cntR[i];
    }
    printf("%lld\n", ans);
    return 0;
}

/*
*/

原文地址：https://www.cnblogs.com/CJLHY/p/10990265.html