UVa 11039 - Building designing
2015-03-11 23:09
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题目:有n个绝对值不为0的数字,从中找到一个序列,正负交替,绝对值递增,求序列最大长度。
分析:dp,动态规划。因为绝对值要递增,所以先按绝对值排序。
设前k个数组成大的序列最长为f(k),则有如下地推关系:
f(k)= f(k-1) { data[k]*data[k-1] > 0,最后量元素不同时取 }
= f(k-1)+ 1 { data[k]*data[k-1] < 0,最后量元素同时取 }
(所有数据均不相同,且不为零)
说明:(⊙v⊙)。
分析:dp,动态规划。因为绝对值要递增,所以先按绝对值排序。
设前k个数组成大的序列最长为f(k),则有如下地推关系:
f(k)= f(k-1) { data[k]*data[k-1] > 0,最后量元素不同时取 }
= f(k-1)+ 1 { data[k]*data[k-1] < 0,最后量元素同时取 }
(所有数据均不相同,且不为零)
说明:(⊙v⊙)。
#include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> #include <cmath> using namespace std; int data[500005],f[500005]; bool cmp(int a, int b) { return abs(a) < abs(b); } int main() { int p,n; while (~scanf("%d",&p)) while (p --) { scanf("%d",&n); for (int i = 0 ; i < n ; ++ i) scanf("%d",&data[i]); sort(data, data+n, cmp); f[0] = 1; if (!n) f[0] = 0; for (int i = 1 ; i < n ; ++ i) if (1.0*data[i]*data[i-1] > 0) f[i] = f[i-1]; else f[i] = f[i-1]+1; printf("%d\n",f[n-1]); } return 0; }
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