hdu 4666 Hyperspace（线段树）

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题目链接：hdu 4666 Hyperspace

解题思路

线段树单点修改，区间查询。做2k维护ans数组即可。

代码

#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\n", ans[i]);    }    return 0;}

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