arc076f F - Exhausted?

ARC076 F - Exhausted?

[题目大意]

\(有m个座位,分别位于坐标为1,2,3,...,m的地方;n个客人,第i位客人只坐位于[0,li]∪[ri,m]的座位。每个座位只能坐一个人,问最少需要添加几个座位才能使所有人坐下?\)

[Solution]

本题考察对霍尔定理的理解, $对于二分图G=<V_1, V_2, E> , 设|V_1| < |V_2| , 则G中存在V_1 到V_2的完美匹配当且仅当对于任意S\subset V_1, 对于S的邻域N(S), 均有|S| \le |N(S)| $

而霍尔定理有一个推论, 就是若使G 中存在完美匹配, 则最少补充\(max\{0, |S| - |N(S)|\}\) 条边

回到本题, 对于一个人, 把他看做左部点, 把座位1到m看做右部, 将客人向所有\(i\in[1, l_i] \cup [r_i, m]\)连边

因为左部S所对应的右部节点的形式为\([1, l] \cup [r, m]\) 节点数为\(m - (r - l + 1)= m - r + l - 1\)

那么依据霍尔定理, 答案就是 \(|S| - m + r - l + 1\)

那么, 求出上述式子的最大值即可, 但是枚举S的复杂度太高, 于是考虑右部区间\([L, R]\) 找出对应的左部节点,所以可以将人\(l_i, r_i\)\(l_i\) 升序, 建一棵线段树存储\(r_i\) ,将右部节点映射上去, 并把初值设为i,。迭代\(l\) , 每次将左端点是\(l\)\(r_i\), 更新入线段树, 即将\([l, r]\) 区间加一, 每次求出\([L, m]\)的最大值。

对于\(r_i = m + 1\)的点, 在额外建一个点m + 1,存储即可

#include <bits/stdc++.h>
#define for_(i,a,b) for (ll i = (a); i < (b); i++)
#define rep_(i,a,b) for (ll i = (a); i <= (b); i++)
#define ll long long
#define pii pair<int, int>
#define fi first
#define se second
#define sz(a) a.size()
#define all(v) v.begin(), v.end()
#define int long long
using namespace std;
const int maxn = 5e5 + 10, mod = 1e9 + 7;// mod = 1949777;
const double EPS = 1e-3;
int n, m;
vector<int> a[maxn];
struct segment_tree{
    int N;
    long long P;
    vector<long long> ST, A, M, F; // A -> tag add  / M -> tag mul /F -> tag max
    segment_tree(int n) {
        N = 1;
        while(N < n) {
            N *= 2;
        }
        ST = vector<long long>(4 * N - 1, 0);
        A = ST;
        M = vector<long long>(4 * N - 1, 1);
        F = A;
        P = 1e15;
    }
    void pushdown(int s, int t, int p) {
        int l = 2 * p, r = 2 * p + 1, mid = s + ((t - s) >> 1);
        if (M[p] != 1) {
            M[l] *= M[p], M[l] %= P;
            A[l] *= M[p], A[l] %= P;
            ST[l] *= M[p], ST[l] %= P;
            F[l] *= M[p], F[l] %= P;
            M[r] *= M[p], M[r] %= P;
            A[r] *= M[p], A[r] %= P;
            ST[r] *= M[p], ST[r] %= P;
            F[r] *= M[p], F[r] %= P;
            M[p] = 1;
        }
        if (A[p] != 0) {
            ST[l] += A[p] * (mid - s + 1), ST[l] %= P;
            A[l] += A[p], A[l] %= P;
            F[l] += A[p], F[l] %= P;
            ST[r] += A[p] * (t - mid), ST[r] %= P;
            A[r] += A[p], A[r] %= P;
            F[r] += A[p], F[r] %= P;
            A[p] = 0; 
        }
        return;
    }
    void pushup(int p) {
        int l = 2 * p, r = 2 * p + 1;
        ST[p] = (ST[l] + ST[r]) % P;
        F[p] = max(F[l], F[r]);
        return;
    }
    void mul(int l, int r, int c, int s, int t, long long p) {
        if (l <= s && t <= r) {
            M[p] *= c, A[p] *= c, ST[p] *= c;
            M[p] %= P, A[p] %= P, ST[p] %= P;
            F[p] *= c, F[p] %= P;
            return;
        }
        int mid = s + ((t - s) >> 1);
        pushdown(s, t, p);
        if (l <= mid) mul(l, r, c, s, mid, p * 2);
        if (mid < r) mul(l, r, c, mid + 1, t, p * 2 + 1);
        //(ST[p] = ST[p * 2] + ST[p * 2 + 1]) %= P;
        pushup(p);
    }
    void add(int l, int r, int c, int s, int t, long long p) {
        if (l <= s && t <= r) {
            ST[p] += (t - s + 1) * c; ST[p] %= P;
            A[p] += c; A[p] %= P;
            F[p] += c, F[p] %= P;
            return;
        }
        int mid = s + ((t - s) >> 1);        
        pushdown(s, t, p);
        if (l <= mid) add(l, r, c, s, mid, 2 * p);
        if (mid < r) add(l, r, c, mid + 1, t, 2 * p + 1);
        //(ST[p] = ST[2 * p] + ST[2 * p + 1]) %= P;
        pushup(p);
        return;
    }
    long long getsum(int l, int r, int s, int t, int p) {
        if (l <= s && t <= r) {
            return ST[p];
        }
        int mid = s + ((t - s) >> 1);
        pushdown(s, t, p);
        long long res = 0;
        if (l <= mid) res += getsum(l, r, s, mid, 2 * p), res %= P;
        if (mid < r) res += getsum(l, r, mid + 1, t, 2 * p + 1), res %= P;
        return res;
    }
    int getmax(int l, int r, int s, int t, int p) {
        if (l <= s && t <= r) return F[p];
        pushdown(s, t, p);
        int mid = s + ((t - s) >> 1);
        int res = -1e15;
        if (l <= mid) res = max(res, getmax(l, r, s, mid, 2 * p));
        if (mid < r) res = max(res, getmax(l, r, mid + 1, t, 2 * p + 1));
        return res; 
    }
}; 
signed main() {
    cin >> n >> m;
    int ans = max(0LL, n - m);
    m++; // 由于线段树板子是Indexed_1,所以坐标整体+1
    segment_tree T(m + 2);
    for (int i = 1; i <= m + 1; i++) T.add(i, i, i - 1, 1, m + 1, 1);
    for (int i = 1; i <= n; i++) {
        int u, v; cin >> u >> v;
        u++, v++; 
        a[u].push_back(v);
    } 
    for (int i = 1; i <= m; i++) {
        for (int j = 0; j < (int)a[i].size(); j++) {
            T.add(1, a[i][j], 1, 1, m + 1, 1);    
        }
        ans = max(ans, -(i - 1) - (m - 1) - 1 + T.getmax(i + 1, m + 1, 1, m + 1, 1));     //即上文的式子  
    }
    cout << ans << endl;
    return 0;
}

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