[USACO 2011 Nov Gold] Above the Median【逆序对】
2016-10-19 20:04
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传送门:http://www.usaco.org/index.php?page=viewproblem2&cpid=91
这一题我很快的想出了,把>= x的值改为1,< x的改为-1,这样,某一段连续区间的和若非负,则这段区间符合题意。我连这个都想到了却没想到要去求逆序对!!!天呐,我真是太迷了!!
前缀和+逆序对。
逆序对就不解释了,该归并的归并,Fenwick Tree的Fenwick Tree,但是对于此题有一个O(n)的算法,仅限于求此题的逆序对!鉴于这一题值域范围小,而且相邻的两个前缀和只会相差1,所以可以针对这个特点设计算法,详见代码(真的不好说...)
#include <cstdio> #include <cstring> const int maxn = 100005, zero = 100000; int n, s, a, x; long long ans, f[maxn << 1], t; int main(void) { freopen("median.in", "r", stdin); freopen("median.out", "w", stdout); scanf("%d%d", &n, &x); ans = (long long)n * (long long)(n + 1) >> 1; f[zero] = 1; for (int i = 1; i <= n; ++i) { scanf("%d", &a); if (a >= x) { ++s; t -= f[s + zero]; } else { t += f[s + zero]; --s; } ++f[s + zero]; ans -= t; } printf("%I64d\n", ans); return 0; }
代码中f[i]表示这之前,i这个值出现了几次。t表示的是以本次的i为结尾的逆序对数。
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