The Tower of Babylon UVA

题目：题目链接

思路：每个方块可以用任意多次，但因为底面限制，每个方块每个放置方式选一个就够了，以x y为底 z 为高，以x z为底 y 为高，以y z为底 x为高，因为数据量很小，完全可以把每一种当成DAG上的一个结点，然后建图找最长路径。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <list>

#define FRER() freopen("in.txt", "r", stdin)
#define FREW() freopen("out.txt", "w", stdout)

#define INF 0x3f3f3f3f

using namespace std;

struct block {
    int x, y, z;
    block() {}
    block(int x, int y, int z):x(x), y(y), z(z) {}
    bool operator < (const block & other) const {
        return (x < other.x && y < other.y) || (x < other.y && y < other.x);
    }
};

vector<block> vec;

int n, m, d[100];
bool G[100][100];

void init() {
    vec.clear();
    int x, y, z;
    for(int i = 0; i < n; ++i) {
        cin >> x >> y >> z;
        vec.push_back(block(x, y, z));
        vec.push_back(block(x, z, y));
        vec.push_back(block(y, z, x));
    }
    m = n * 3;
    memset(G, 0, sizeof(G));
    for(int i = 0; i < m; ++i) 
        for(int j = 0; j < m; ++j)
            if(vec[i] < vec[j])
                G[i][j] = 1;

    memset(d, 0, sizeof(d));
}

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


int solve() {
    int ans = -1;
    for(int i = 0; i < m; ++i) 
        ans = max(ans, dp(i, vec[i].z));
    return ans;
}

int main()
{
    //FRER();
    //FREW();
    ios::sync_with_stdio(0);
    cin.tie(0);
    
    int kase = 0;
    while(cin >> n, n) {
        init();
        cout << "Case " << ++kase << ": maximum height = " << solve() << endl;
    }

    return 0;
}