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; }