DP(DAG) UVA 437 The Tower of Babylon

 

题目传送门

题意：给出一些砖头的长宽高，砖头能叠在另一块上要求它的长宽都小于下面的转头的长宽，问叠起来最高能有多高

分析：设一个砖头的长宽高为x, y, z，那么想当于多了x, z, y 和y, x, z的砖头，如果i能叠在j上，那么g[i][j] = true，转换成DAG问题，dp[i]表示第i块叠在最上部最高的高度

收获：转换成经典模型

 

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-28 18:00:01
* File Name     :UVA_437.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e2 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Block	{
	int x, y, z;
}b[N];
bool g[N][N];
int dp[N];
int n;

int DFS(int u)	{
	if (dp[u] != -1)	return dp[u];
    dp[u] = b[u].z;
	for (int i=1; i<=n; ++i)	{
		if (g[u][i])	{
			dp[u] = max (dp[u], DFS (i) + b[u].z);
		}
	}
	return dp[u];
}

bool check(int i, int j)	{
	if (b[i].x < b[j].x && b[i].y < b[j].y)	return true;
	if (b[i].x < b[j].y && b[i].y < b[j].x)	return true;
	return false;
}

int main(void)    {
    int cas = 0;
	while (scanf ("%d", &n) == 1)	{
		if (n == 0)	break;
		for (int i=1; i<=n; ++i)	{
			scanf ("%d%d%d", &b[i].x, &b[i].y, &b[i].z);
			b[n+i].x = b[i].x, b[n+i].y = b[i].z, b[n+i].z = b[i].y;
			b[2*n+i].x = b[i].y, b[2*n+i].y = b[i].z, b[2*n+i].z = b[i].x;
		}
		memset (g, false, sizeof (g));
		n *= 3;
		for (int i=1; i<=n; ++i)	{
			for (int j=i+1; j<=n; ++j)	{
				if (check (i, j))	g[i][j] = true;
				if (check (j, i))	g[j][i] = true;
			}
		}
		memset (dp, -1, sizeof (dp));
		int ans = 0;
		for (int i=1; i<=n; ++i)	{
			ans = max (ans, DFS (i));
		}
		printf ("Case %d: maximum height = %d\n", ++cas, ans);
	}

    return 0;
}