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  • BZOJ3524: [Poi2014]Couriers(主席树)

    题意

    题目链接

    Sol

    严格众数只会出现一次,那么建出主席树,维护子树siz,直接在树上二分即可

    #include<bits/stdc++.h> 
    #define LL long long 
    using namespace std;
    const int MAXN = 2e6 + 10;
    inline int read() {
        char c = getchar(); int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int N, M, a[MAXN], date[MAXN], rt[MAXN], num;
    //struct GZYAKIOI {
    #define MAX (MAXN * 10)
        int tot, ls[MAX], rs[MAX], siz[MAX];
        void insert(int &rt, int pre, int l, int r, int v) {
            rt = ++tot; ls[rt] = ls[pre]; rs[rt] = rs[pre]; siz[rt] = siz[pre] + 1;
            if(l == r) return ;
            int mid = l + r >> 1;
            if(v <= mid) insert(ls[rt], ls[pre], l, mid, v);
            else insert(rs[rt], rs[pre], mid + 1, r, v);
        }
        int Query(int pre, int rt, int l, int r, int v) {
            if(l == r) return (siz[rt] - siz[pre] > v ? date[l] : 0);  
            int mid = l + r >> 1;
            if(siz[ls[rt]] - siz[ls[pre]] > v) return Query(ls[pre], ls[rt], l, mid, v);
            else return Query(rs[pre], rs[rt], mid + 1, r, v);
        }
    //}T;
    
    signed main() {
        N = read(); M = read();
        for(int i = 1; i <= N; i++) a[i] = date[i] = read();
        sort(date + 1, date + N + 1);
        num = unique(date + 1, date + N + 1) - date - 1;
        for(int i = 1; i <= N; i++) a[i] = lower_bound(date + 1, date + num + 1, a[i]) - date, insert(rt[i], rt[i - 1], 1, num, a[i]); 
        while(M--) {
            int l = read(), r = read();
            printf("%d
    ", Query(rt[l - 1], rt[r], 1, num, (r - l + 1) / 2));
        }
        return 0;
    }
    /*
    7 5
    1 1 3 2 3 4 3
    6 6
    1 3
    1 4
    3 7
    1 7
    */
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/10213388.html
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