UVA11039- Building designing
2014-08-06 21:33
197 查看
题意:有n个绝对值各不相同的非0整数,选出尽量多的数,排成一个序列,使得正负号交替,且绝对值递增。输出最长序列长度。
思路:其实按照绝对值排序后,只要选出正负号交替最长的序列就可以了。用一个标记来表示下一个要选的是正号还是负号。
思路:其实按照绝对值排序后,只要选出正负号交替最长的序列就可以了。用一个标记来表示下一个要选的是正号还是负号。
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; const int MAXN = 500005; struct node{ int x, y; }a[MAXN]; int arr[MAXN], b[MAXN], order[MAXN]; int n; int cmp(node a, node b) { return a.y < b.y; } int main() { int cas; scanf("%d", &cas); while (cas--) { scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%d", &a[i].x); if (a[i].x < 0) a[i].y = abs(a[i].x); else a[i].y = a[i].x; } sort(a, a + n, cmp); int cnt = 1, flag ; if (a[0].x == a[0].y) flag = 1; else flag = 0; for (int i = 1; i < n; i++) { if (a[i].x == a[i].y && !flag) { flag = 1; cnt++; } else if (a[i].x != a[i].y && flag) { flag = 0; cnt++; } } printf("%d\n", cnt); } return 0; }
相关文章推荐
- UVa 11039 Building designing (求按绝对值升序最长正负交替数列长度)
- UVA 11039
- UVA 11039 Building designing
- UVA 11039 - Building designing
- UVa 11039 - Building designing
- UVA11039 Building designing
- UVa 11039 设计建筑物
- UVA 11039 Building designing
- UVA 11039(p78)----Building designing
- uva11039 Building designing
- UVA11039 - Building designing
- uva 11039 - Building designing(贪心)
- UVa 11039 - Building designing
- 贪心水题。UVA 11636 Hello World,LA 3602 DNA Consensus String,UVA 10970 Big Chocolate,UVA 10340 All in All,UVA 11039 Building Designing
- UVA 11039 Building designing
- UVA11039 Building designing【排序】
- Uva11039 Building design...
- UVA - 11039 Building designing (sort)
- UVA - 11039 Building designing
- UVa 11039 - Building Designing