ACdream 1429 Rectangular Polygon

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Problem Description

A rectangular polygon is a polygon whose edges are all parallel to the coordinate axes. The polygon must have a single, non-intersecting boundary. No two adjacent sides must be parallel.

Johnny has several sticks of various lengths. He would like to construct a rectangular polygon. He is planning to use sticks as horizontal edges of the polygon, and draw vertical edges with a pen.

Now Johnny wonders, how many sticks he can use. Help him, find the maximal number of sticks that Johnny can use. He will use sticks only as horizontal edges.

Input

The first line of the input file contains n — the number of sticks (1 ≤ n ≤ 100). The second line contains n integer numbers — the lengths of the sticks he has. The length of each stick doesn’t exceed 200.

Output

Print l — the number of sticks Johnny can use at the first line of the output file. The following 2l lines must contain the vertices of the rectangular polygon Johnny can construct. Vertices must be listed in order of traversal. The first two vertices must be the ends of a horizontal edge. If there are several solution, output any one. Vertex coordinates must not exceed 109
. If no polygon can be constructed, output l = 0.

Sample Input

41 2 3 541 2 4 841 1 1 1

Sample Output

30 01 01 13 13 20 2040 01 01 12 12 -21 -21 -10 -1

Hint

单组数据

In the first example Johnny uses a stick of length 1 for (0, 0)−(1, 0) edge, a stick of length 2 for (1, 1)−(3, 1) edge and a stick of length 3 for (3, 2) − (0, 2) edge. There is no way to use all four sticks.

Source

Andrew Stankevich Contest 23

Manager

mathlover

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动规~

做了两天的题，思路不难，但是写起来很麻烦……而且最后还要求输出组成最大值的边的位置，一脸懵啊……方法还真是神奇，用了一个fa数组记录前继边的长度，最后用l和r递推输出~

其实题目就相当于是用小木棍拼成两条长度相等的边使得它们最长，用f[i][j]表示已用i根木棍，上下差值为j的最大长度，C++要再加上20000（100*200）防止下表出现负数。

数组开小了循环总是莫名其妙地终止……后来才发现真是纠结啊……

#include<cstdio>#include<cstring>int n,a[101],f[101][40001],ans1[101],ans2[101],l,r,fa[101][40001],now;int main(){while(scanf("%d",&n)==1){memset(f,-1,sizeof(f));f[0][20000]=0;for(int i=1;i<=n;i++) scanf("%d",&a[i]);l=r=20000;for(int i=1;i<=n;i++){for(int j=l;j<=r;j++){if(f[i-1][j]<=-1) continue;if(f[i][j]<f[i-1][j]){f[i][j]=f[i-1][j];fa[i][j]=j;}if(f[i][j+a[i]]<f[i-1][j]+1){f[i][j+a[i]]=f[i-1][j]+1;fa[i][j+a[i]]=j;}if(f[i][j-a[i]]<f[i-1][j]+1){f[i][j-a[i]]=f[i-1][j]+1;fa[i][j-a[i]]=j;}}l-=a[i];r+=a[i];}printf("%d\n",f[n][20000]);now=20000;ans1[0]=ans2[0]=0;for(int i=n;i>=1;i--){int k=fa[i][now];if(k<now) ans1[++ans1[0]]=now-k;if(k>now) ans2[++ans2[0]]=k-now;now=k;}l=r=0;for(int i=1;i<=ans2[0];i++){r++;printf("%d %d\n",l,r);l+=ans2[i];printf("%d %d\n",l,r);}r=0;for(int i=1;i<=ans1[0];i++){r--;printf("%d %d\n",l,r);l-=ans1[i];printf("%d %d\n",l,r);}}return 0;}

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