POJ 2241 The Tower of Babylon

标签：poj   algorithm   dp   

题目地址:

http://poj.org/problem?id=2241

题意:

给定N中方块，给定他们的长宽高，然后每一种方块的个数都为无限个。将任意数目的方块叠起来（但是要满足叠在一起的方块中，上面的方块的长和宽必须都小于下面方块的长和宽），问这些方块最多能叠多高。

思路:

DP求解。对Blocks的先按照X降级，再按照Y降级排序，可转化为最长公共子序列问题，即求子序列权值之和最大。可化为图算法，较笨。

#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;

#define MAX_SIZE 300

struct Block{
    int x;
    int y;
    int height;
};

int nums[3];
Block blocks[MAX_SIZE];

bool cmp( Block a, Block b ){
    if( a.x == b.x )
        return a.y > b.y;
    return a.x > b.x;
}

int main()
{
    int case_num = 0;
    while( true ){
        case_num += 1;
        int num;
        cin >> num;
        if( num == 0 ) break;
        int index = 0;
        for( int i = 0; i < num; ++i ){
            cin >> nums[0] >> nums[1] >> nums[2];
    
            int x = max( max( nums[0], nums[1] ), nums[2] );
            int z = min( min( nums[0], nums[1] ), nums[2] );
            int y = nums[0] + nums[1] + nums[2] - x - z;

            blocks[index].x = x;
            blocks[index].y = y;
            blocks[index].height = z;
            index++;
            blocks[index].x = x;
            blocks[index].y = z;
            blocks[index].height = y;
            index++;
            blocks[index].x = y;
            blocks[index].y = z;
            blocks[index].height = x;
            index++;
        }
        sort( blocks, blocks + index, cmp );
        for( int i = 0; i < index; ++i ){
            int max_height = 0;
            for( int j = 0; j < i; ++j ){
                if( blocks[j].y > blocks[i].y && blocks[j].x > blocks[i].x ){
                    max_height = max( max_height, blocks[j].height );
                }
            }
            blocks[i].height += max_height;
        }
        int max_height = 0;
        for( int i = 0; i < index; ++i ){
            max_height = max( max_height, blocks[i].height );
        }
        cout << "Case " << case_num <<": maximum height = " << max_height << endl;
    }
    return 0;
}

POJ 2241 The Tower of Babylon,布布扣,bubuko.com

POJ 2241 The Tower of Babylon

标签：poj   algorithm   dp   

原文地址：http://blog.csdn.net/u011659057/article/details/27570551