HDU

题目大意：给出一系列方块，要求上面的方块长宽都比下面的小，问最高能叠多高。每个方块可以翻转，并且个数不限。
解题思路：可以翻转意味着一组长宽高能有 6 种摆放方式，列出所有方式按照长宽排序，dp记录当前方块之前能够叠的最高高度。

#include<iostream>#include<stdio.h>#include<algorithm>#include<cmath>#include<string.h>#include<string>#include<queue>#include<map>#define max(a,b) ((a)>(b)?(a):(b))#define min(a,b) ((a)<(b)?(a):(b))const int INF = 0x3f3f3f3f;const int NINF = -INF -1;const int MAXN = 180+5;using namespace std;struct point {    int x, y, z;};point p[MAXN];int dp[MAXN];int n, ans, cnt = 0;bool cmp(point a, point b) {    if (a.x == b.x)        return a.y < b.y;    return a.x < b.x;}int main() {    while (scanf("%d", &n) && n) {        memset(dp, 0, sizeof(dp));        int t = 0;        int x, y, z;        for (int i = 0; i < n; i++) {            scanf("%d%d%d", &x, &y, &z);            p[t].x = x, p[t].y = y, p[t].z = z, t++;            p[t].x = x, p[t].y = z, p[t].z = y, t++;            p[t].x = y, p[t].y = x, p[t].z = z, t++;            p[t].x = y, p[t].y = z, p[t].z = x, t++;            p[t].x = z, p[t].y = x, p[t].z = y, t++;            p[t].x = z, p[t].y = y, p[t].z = x, t++;        }        sort(p, p+t, cmp);//      for (int i = 0; i < t; i++)//          printf("%d %d %d\n", p[i].x, p[i].y, p[i].z);        dp[0] = p[0].z;        ans = 0;        for (int i = 1; i < t; i++) {            dp[i] = p[i].z;            for (int j = 0; j < i; j++)                if (p[j].x < p[i].x && p[j].y < p[i].y)                    dp[i] = max(dp[j]+p[i].z, dp[i]);//          printf("%d %d\n", i, dp[i]);            if (dp[i] > ans) ans = dp[i];        }        printf("Case %d: maximum height = %d\n", ++cnt, ans);    }    return 0;}