fzu 2136 取糖果(線段樹)

fzu 2136 取糖果(線段樹)

題目鏈接：fzu 2136 2136 取糖果

題目大意：略。

解題思路：線段樹區間合並。將袋子按照個數排序，每次將最小的放入線段樹，如果當前連續的個數超過區間，那麼說

明最小值即為最後加入的袋子糖果個數。

#include 
#include 
#include 
#include 

using namespace std;

const int maxn = 1e5 + 5;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], L[maxn << 2], R[maxn << 2], S[maxn << 2];

void pushup(int u) {
    S[u] = max(max(S[lson(u)], S[rson(u)]), R[lson(u)] + L[rson(u)]);
    L[u] = L[lson(u)] + (L[lson(u)] == rc[lson(u)] - lc[lson(u)] + 1 ? L[rson(u)] : 0);
    R[u] = R[rson(u)] + (R[rson(u)] == rc[rson(u)] - lc[rson(u)] + 1 ? R[lson(u)] : 0); 
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    L[u] = R[u] = S[u] = 0;

    if (l == r)
        return;

    int mid = (l + r) >> 1;
    build(lson(u), l,  mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

void modify(int u, int x, int w) {
    if (lc[u] == x && x == rc[u]) {
        S[u] = R[u] = L[u] = w ? rc[u] - lc[u] + 1 : 0;
        return;
    }

    int mid = (lc[u] + rc[u]) >> 1;
    if (x <= mid)
        modify(lson(u), x, w);
    else
        modify(rson(u), x, w);
    pushup(u);
}

typedef pair pii;
vector vec;

int main () {
    int cas;
    scanf("%d", &cas);

    while (cas--) {
        int n, x;
        vec.clear();

        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &x);
            vec.push_back(make_pair(x, i));
        }

        sort(vec.begin(), vec.end());

        build(1, 1, n);
        int mv = 0;
        for (int k = 1; k <= n; k++) {
            while (S[1] < k)
                modify(1, vec[mv++].second, 1);
            printf("%d\n", vec[mv-1].first);
        }
    }
    return 0;
}