UVA - 437（DAG模型）

一个盒子的高有三种状态，只要比较长和宽就行了。n个正方体可以看成n×3个矩形嵌套问题。可以转化成有向无环图，建完图后用记忆化搜索即可。

状态转移方程：d(i) = max{d(j) + box[i].z}

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 30 * 3 + 10;

struct Box {
    int x;
    int y;
    int z;
} box[maxn];

int d[maxn];
bool G[maxn][maxn];
int n;

int dp(int i) {
    int& ans = d[i];
    if (ans > 0) return ans;
    ans = box[i].z;
    for(int j = 1; j <= 3*n; j++) {
        if (G[i][j]) ans = max(ans, dp(j) + box[i].z);
    }
    return ans;
}

bool judge(Box a, Box b) {    // 判断a能否连到b
    return (a.x > b.x && a.y > b.y || a.x > b.y && a.y > b.x);
}

int main() {
    int kase = 0;
    freopen("input.txt", "r", stdin);
    while(~scanf("%d", &n) && n) {
        memset(G, false, sizeof(G));
        memset(d, -1, sizeof(d));
        for(int i = 1; i < 3 * n; i += 3) {
            scanf("%d%d%d", &box[i].x, &box[i].y, &box[i].z);
            box[i+1].x = box[i].z;
            box[i+1].y = box[i].x;
            box[i+1].z = box[i].y;
            box[i+2].x = box[i].y;
            box[i+2].y = box[i].z;
            box[i+2].z = box[i].x;
        }
        for(int i = 1; i <= 3 * n; i++) {
            for(int j = 1; j <= 3 * n; j++) {
                if (i == j) continue;
                else {
                    if (judge(box[i], box[j])) {
                        G[i][j] = true;
                    }
                }
            }
        }
        int ans = -0x7fffffff;
        for(int i = 1; i <= 3 * n; i++) {
            ans = max(ans, dp(i));
        }
        printf("Case %d: maximum height = %d\n", ++kase, ans);
    }
    return 0;
}