HDU 6047 Maximum Sequence 贪心 区间最值

链接

http://acm.hdu.edu.cn/showproblem.php?pid=6047

题意

给两个长度为n的序列a, b，现要求max(sum(a[n + 1] + … + a[2n]))，其中a[i] = max(a[b[k]] … a[i - 1])，其中b[k]为b[1 … n]中的一个，每个a[i]选择的b[k]不能重复

思路

写这个博客蹭点访问量2333

肯定先选能选到的最大的数，贪心就好，用一个线段树维护区间最值

代码

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;template<class T1, class T2> inline void gmax(T1& a, T2 b) { if (a < b) a = b; }typedef long long LL;const int MAXN = 3e5 + 5;const int MAXM = 1 << 21;const int MOD = 1e9 + 7;int n, m;int seg_tree[MAXM], BASE;int a[MAXN << 1], b[MAXN];void update(int x) {  x += BASE;  seg_tree[x] = a[x - BASE] - x + BASE;  for (x >>= 1; x; x >>= 1) seg_tree[x] = max(seg_tree[x << 1], seg_tree[x << 1 | 1]);}int query(int l, int r) {  int ret = 0;  l += BASE - 1; r += BASE + 1;  for (; l ^ r ^ 1; l >>= 1, r >>= 1) {    if (~ l & 1) gmax(ret, seg_tree[l ^ 1]);    if (r & 1) gmax(ret, seg_tree[r ^ 1]);  }  return ret;}int main() {  while (~scanf("%d", &n)) {    for (BASE = 1; BASE <= 2 * n + 1; BASE <<= 1);    for (int i = 0; i <= BASE << 1; ++i) seg_tree[i] = 0;    for (int i = 1; i <= n; ++i) {      scanf("%d", a + i);      update(i);    }    for (int i = 0; i < n; ++i) scanf("%d", b + i);    sort(b, b + n);    LL ans = 0;    for (int i = 0; i < n; ++i) {      a[i + n + 1] = query(b[i], i + n);      update(i + n + 1);      (ans += a[i + n + 1]) %= MOD;    }    printf("%I64d\n", ans);  }}