HDU 5813 Elegant Construction(优雅建造) Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Description 题目描述 Being an ACMer requires knowledge in many fields, because problems in this contest may use physics, biology, and…

Elegant Construction 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5813 Description Being an ACMer requires knowledge in many fields, because problems in this contest may use physics, biology, and even musicology as background. And now in this prob…

Elegant Construction 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5813 Description Being an ACMer requires knowledge in many fields, because problems in this contest may use physics, biology, and even musicology as background. And now in this prob…

构造.从a[i]最小的开始放置,例如放置了a[p],那么还未放置的,还需要建边的那个点 需求量-1,然后把边连起来. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<…

可以直接见这个博客:http://blog.csdn.net/black_miracle/article/details/52164974. 对其中的几点作一些解释: 1.这个方法我们对队列中取出的元素,把仍有出度的点连接到这个点时,这个点是不会连接到多余的点的,因为:一个点出队列的时候其他没入队的点都已经连接到这个点过了,后来再一个点出队列的时候,我们拿此时没入队的点连接到它时,肯定已经连接过之前的点了,所以肯定不会增加到达的点数.(可能我也用语言讲不太清楚233...) 2.为什么这个人的博…

Elegant Construction                                                                         Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)                                                                          …

题目链接: Elegant Construction Time Limit: 4000/2000 MS (Java/Others)     Memory Limit: 65536/65536 K (Java/Others) Problem Description Being an ACMer requires knowledge in many fields, because problems in this contest may use physics, biology, and even…

Being an ACMer requires knowledge in many fields, because problems in this contest may use physics, biology, and even musicology as background. And now in this problem, you are being a city architect! A city with N towns (numbered 1 through N) is und…

题意:给定一些点(xi,yi)(xj,yj)满足:i<j,xi<xj,yi>yj.用下面的连起来,使得所有边的长度最小? 思路:考虑用区间表示,f[i][j]表示将i到j的点连起来的最小代价. 那么f[i][j]=min(f[i][k]+f[k+1][j]+cost(i,j) cost(i,j)=a[k].y-a[j].y+a[k+1].x-a[i].x; 看起来和四边形不等式有关系,我们需要证明以下(a<b<c<d) cost(a,c)+cost(b,d)<=c…

题目链接 贴个教程: 四边形不等式学习笔记 \(Description\) 给出平面上的\(n\)个点,满足\(X_i\)严格单增,\(Y_i\)严格单减.以\(x\)轴和\(y\)轴正方向作边,使这\(n\)个点构成一棵树,最小化树边边的总长. \(Solution\) 考虑有两棵构造好的树,要合并这两棵树,要从右边的树中找一个最优点连到左边的树上 不难想到区间DP(真的想不到==) \(f[i][j]\)表示将\([i,j]\)合并为一棵树的最小代价,那么有 \(f[i][j] = \min…