uva 11039 - Building designing
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思路:贪心(排序+枚举)
分析:
1 题目要求按照题目给定的数据求能够建立最高几层的楼房
2 题目要求上层的大小要比下层小,并且两层之间的颜色要交替的出现。很明显,我们只要按照size进行排序,然后枚举即可。
代码:
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int MAXN = 500010;int n , Case;struct floor{ int color; int size; bool operator<(const floor &fl)const{ return size > fl.size; }};floor f[MAXN];int main(){ scanf("%d" , &Case); while(Case--){ scanf("%d" , &n); for(int i = 0 ; i < n ; i++){ scanf("%d" , &f[i].color); f[i].size = abs(f[i].color); } sort(f , f+n); int ans , preSize , preColor; ans = 1; preSize = f[0].size; preColor = f[0].color; for(int i = 1 ; i < n ; i++){ if(f[i].color > 0){ if(preColor < 0){ ans++; preColor = f[i].color; } } else{ if(preColor > 0){ ans++; preColor = f[i].color; } } } printf("%d\n" , ans); } return 0;}
- UVA 11039 Building designing
- UVA 11039 - Building designing
- uva 11039 - Building designing
- Uva-11039-Building designing
- UVa 11039 - Building designing
- Uva - 11039 - Building designing
- UVA 11039 - Building designing
- uva 11039 Building designing
- Uva 11039 - Building designing
- UVA 11039 - Building designing
- UVA - 11039 Building designing
- UVA 11039 Building designing
- UVA 11039 Building designing
- UVA - 11039 Building designing
- UVa 11039 - Building Designing
- UVa 11039 - Building designing
- UVA - 11039 Building designing
- UVA 11039 Building Designing
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