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Codeforces 85D Sum of Medians(线段树)

题目链接:Codeforces 85D - Sum of Medians

题目大意:N个操作,add x:向集合中添加x;del x:删除集合中的x;sum:将集合排序后,将集合中所有下标i % 5 = 3的元素累加求和。

解题思路:线段树单点更新,每个点维护5个值,分别表示从该段区间中i % 5 = t的和。然后两端区间合并时只需要根据左孩子中元素的个数合并。所以有一个c表示区间上元素的个数。因为有相同的数,所以要离线操做,将所有的数映射成位置,但是对于del则不需要映射,因为集合中肯定有才能减掉。那么add和sum操作都是可以搞定了,只剩下del操作,对于del x,x肯定在集合中出现过,所以每次删除第一个x即可,如果快速查找,要借助map和一个辅助数组,因为删除一个后要重新映射,所以借助辅助数组。

#include <cstdio>
#include <cstring>
#include <map>
#include <vector>
#include <algorithm>

using namespace std;

typedef long long ll;
const int mod = 5;
const int maxn = 1e5+5;

int N, M, pos[maxn], v[maxn];
map<ll, int> G;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], c[maxn << 2];
ll s[maxn << 2][6];

inline void maintain (int u, int d) {
    c[u] += d;
    memset(s[u], 0, sizeof(s[u]));
    s[u][0] = (c[u] ? pos[lc[u]] : 0);
}

inline void pushup(int u) {
    int t = c[lson(u)] % mod;
    c[u] = c[lson(u)] + c[rson(u)];
    for (int i = 0; i < mod; i++)
        s[u][i] = s[lson(u)][i] + s[rson(u)][(i + mod - t) % mod];
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    c[u] = 0;
    memset(s[u], 0, sizeof(s[u]));

    if (l == r)
        return;
    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

void modify (int u, int x, int d) {
    if (lc[u] == x && rc[u] == x) {
        maintain(u, d);
        return;
    }

    int mid = (lc[u] + rc[u]) / 2;
    if (x <= mid)
        modify(lson(u), x, d);
    else
        modify(rson(u), x, d);
    pushup(u);
}

struct OP {
    int p, k, id;
    OP (int k = 0, int p = 0, int id = 0) {
        this->k = k;
        this->p = p;
        this->id = id;
    }
    friend bool operator < (const OP& a, const OP& b) {
        if (a.k == 0)
            return false;
        if (b.k == 0)
            return true;
        if (a.p != b.p)
            return a.p < b.p;
        return a.id < b.id;
    }
};

inline bool cmp (const OP& a, const OP& b) {
    return a.id < b.id;
}

vector<OP> vec;

void init () {
    scanf("%d", &N);
    char op[5];
    int x;

    for (int i = 1; i <= N; i++) {
        scanf("%s", op);
        if (op[0] == ‘s‘)
            vec.push_back(OP(0, 0, i));
        else {
            scanf("%d", &x);
            vec.push_back(OP(op[0] == ‘a‘ ? 1 : -1, x, i));
        }
    }

    M = 1;
    sort(vec.begin(), vec.end());
    for (int i = 0; i < N; i++) {
        if (vec[i].k < 0)
            continue;
        if (vec[i].k == 0)
            break;

        pos[M] = vec[i].p;
        vec[i].p = M++;
    }
    build(1, 1, M);
}

void solve () {
    sort(vec.begin(), vec.end(), cmp);
    for (int i = 0; i < N; i++) {
        //printf("%d %d!\n", vec[i].k, pos[vec[i].p]);
        if (vec[i].k == 0)
            printf("%lld\n", s[1][2]);
        else if (vec[i].k == -1) {
            int tmp = vec[i].p;
            v[G[tmp]] = 0;
            modify(1, G[tmp], -1);

            if (G[tmp] <= N && v[G[tmp]+1] && pos[G[tmp]] == pos[G[tmp]+1])
                G[tmp]++;
            else
                G[tmp] = 0;
        } else {
            int tmp = pos[vec[i].p];
            v[vec[i].p] = 1;
            modify(1, vec[i].p, 1);
            if (G[tmp] == 0)
                G[tmp] = vec[i].p;
        }
    }
}

int main () {
    init();
    solve();
    return 0;
}

Codeforces 85D Sum of Medians(线段树)