ACdream 1429 Rectangular Polygon

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 10 ⁹
.      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.

dp+记录路径

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int maxn = 40005;int f[105][maxn], p[105][maxn], a[maxn], l, r, n, b[2][maxn];int main(){int i, j;while (~scanf("%d", &n)){memset(f, 0, sizeof(f));memset(p, 0, sizeof(p));f[0][l = r = 20000] = 1;for (i = 1; i <= n; i++){scanf("%d", &a[i]);for (j = l; j <= r; j++){if (f[i][j]<f[i - 1][j]){f[i][j] = f[i - 1][j];p[i][j] = p[i - 1][j];}if (f[i - 1][j]){if (f[i][j + a[i]]<f[i - 1][j] + 1){f[i][j + a[i]] = f[i - 1][j] + 1;p[i][j + a[i]] = -i;}if (f[i][j - a[i]]<f[i - 1][j] + 1){f[i][j - a[i]] = f[i - 1][j] + 1;p[i][j - a[i]] = i;}l = min(l, j - a[i]);r = max(r, j + a[i]);}}}printf("%d\n", f[n][20000] - 1);if (f[n][20000]>1){int k1 = 0, k2 = 0;for (i = p[n][20000], j = 20000; i;){if (i>0){b[1][k2++] = a[i];j += a[i];i = p[i - 1][j];}else{i = -i;b[0][k1++] = a[i];j -= a[i];i = p[i - 1][j];}}int x = 0, y = 0;for (i = 0; i<k1; i++){printf("%d %d\n", x, y);printf("%d %d\n", (x += b[0][i]), y++);}for (i = 0; i<k2; i++){printf("%d %d\n", x, y);printf("%d %d\n", (x -= b[1][i]), y++);}}}return 0;}