uva 437(最长递减子序列)

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题意：给出有n种长方体的长宽高，假如有无数个这些长方体来摆放，要求如果一个长方体的长和宽都比下面的长方体小，那么就可以垒上去，问最多能垒多高。

题解：将所给出的某种长方体的所有高不同的情况都存储起来，然后按面积降序排列，按最长递减子序列的思想，将某个点当做最高点，更新最高的高度。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 35;struct Block {int x, y, z;}block[N * 3];int f[N], n, t = 1, a, b, c;int cmp(Block a, Block b) {return  a.x * a.y > b.x * b.y;}int main() {while (scanf("%d", &n) && n) {int k = 0;for (int i = 0; i < n; i++) {scanf("%d%d%d", &a, &b, &c);block[k].x = a;block[k].y = b;block[k++].z = c;block[k].x = a;block[k].y = c;block[k++].z = b;block[k].x = b;block[k].y = c;block[k++].z = a;}sort(block, block + k, cmp);int res = -1;for (int i = 0; i < k; i++) {f[i] = block[i].z;for (int j = 0; j < i; j++) {if (block[j].x > block[i].x && block[j].y > block[i].y ||block[j].x > block[i].y && block[j].y > block[i].x)f[i] = f[i] > f[j] + block[i].z ? f[i] : f[j] + block[i].z;}if (f[i] > res)res = f[i];}printf("Case %d: maximum height = %d\n", t++, res);}return 0;}

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