HDu 4666 Hyperspace(セグメントツリー)

タイトルリンク:hdu 4666 Hyperspace

問題を解く構想.

セグメントツリーの単点修正、区間クエリー.2 kをしてans配列を維持すればいいです.

コード#コード#

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
const int maxn = 60005;
const int inf = 0x3f3f3f3f;

int lc[maxn<<2], rc[maxn<<2], mi[maxn<<2], mx[maxn<<2];

void pushup(int u) {
    mi[u] = min(mi[lson(u)], mi[rson(u)]);
    mx[u] = max(mx[lson(u)], mx[rson(u)]);
}

void build (int u, int l, int r) {
    lc[u] = l, rc[u] = r;
    mi[u] = inf, mx[u] = -inf;

    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 v) {
    if (lc[u] == rc[u]) {
        mi[u] = mx[u] = v;
        return ;
    }

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

void clear(int u, int x) {
    if (lc[u] == rc[u]) {
        mi[u] = inf;
        mx[u] = -inf;
        return ;
    }

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

int N, K, ans[maxn];
int op[maxn], id[maxn], P[maxn][5];

void init () {
    for (int i = 1; i <= N; i++) {
        scanf("%d", &op[i]);
        if (op[i]) scanf("%d", &id[i]);
        else {
            for (int j = 0; j < K; j++)
                scanf("%d", &P[i][j]);
        }
    }
    memset(ans, 0, sizeof(ans));
}

int get(int* x, int s) {
    int ret = 0;
    for (int i = 0; i < K; i++) {
        if (s&(1<<i)) ret += x[i];
        else ret -= x[i];
    }
    return ret;
}

int main () {
    while (scanf("%d%d", &N, &K) == 2) {
        init();
        for (int i = 0; i < (1<<K); i++) {
            build(1, 1, N);
            for (int j = 1; j <= N; j++) {
                if (op[j]) clear(1, id[j]);
                else {
                    int tmp = get(P[j], i);
                    modify(1, j, tmp);
                }
                ans[j] = max(ans[j], mx[1] - mi[1]);
            }
        }
        for (int i = 1; i <= N; i++)
            printf("%d
", ans[i]);
    }
    return 0;
}