pku 2082 Terrible Sets

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这道题感觉就是在描述几何问题 (largest rectangle )，当然是用代数方式。

第一个想到的方法是DP，O(n^2)，同枚举没什么区别。

再一个容易想到的办法，按高度，最高的有机会“自成一家”，otherwise, the taller part is droped, the remaining is added to taller one on it's sides, and so one. Time complexity: O(nlgn)[sort]+O(n)[merge] = O(nlgn). max N is 50000, worth a try.

The O(n) algorithm given by report: from left to right calculate the area of one higher than both sides, then merge with higher side, and go on.

#include <iostream>#define N 50001using namespace std;struct Rect{int w,h;};Rect rect[N];int top; // use it as a stackint main(){int n;__int64 s;Rect tmp,tmp2;while (scanf("%d",&n)!=EOF){if (n==-1)break;rect[0].h = rect[0].w = 0;s = 0; top =1;for (int i=1; i<=n; ++i){scanf("%d%d",&tmp.w,&tmp.h);while (tmp.h < rect[top-1].h){if (rect[top-1].h*rect[top-1].w>s)                    s = rect[top-1].h*rect[top-1].w;if (rect[top-2].h > tmp.h){rect[top-2].w +=rect[top-1].w;--top;}else{tmp.w += rect[top-1].w;--top;break;}}rect[top] = tmp;++top;}while (top>=2){if (rect[top-1].h*rect[top-1].w>s)                    s = rect[top-1].h*rect[top-1].w;rect[top-2].w += rect[top-1].w;--top;}printf("%I64d/n",s);}}

http://acm.pku.edu.cn/JudgeOnline/problem?id=1755 这道题用另一方式描述了一道计算几何问题。

总的说来是图形化的描述。

做这题时想到了LRJ黑书上的打造王冠那题。