HDU4001-To Miss Our Children Time

To Miss Our Children Time

                                                                                   Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
                                                                                                    Total Submission(s): 4677    Accepted Submission(s): 1297

Problem Description

Do you remember our children time? When we are children, we are interesting in almost everything around ourselves. A little thing or a simple game will brings us lots of happy time! LLL is a nostalgic boy, now he grows up. In the dead of night, he often misses something, including a simple game which brings him much happy when he was child. Here are the game rules: There lies many blocks on the ground, little LLL wants build "Skyscraper" using these blocks. There are three kinds of blocks signed by an integer d. We describe each block's shape is Cuboid using four integers ai, bi, ci, di. ai, bi are two edges of the block one of them is length the other is width. ci is 
thickness of the block. We know that the ci must be vertical with earth ground. di describe the kind of the block. When di = 0 the block's length and width must be more or equal to the block's length and width which lies under the block. When di = 1 the block's length and width must be more or equal to the block's length which lies under the block and width and the block's area must be more than the block's area which lies under the block. When di = 2 the block length and width must be more than the block's length and width which lies under the block. Here are some blocks. Can you know what's the highest "Skyscraper" can be build using these blocks?

 

Input

The input has many test cases. 
For each test case the first line is a integer n ( 0< n <= 1000) , the number of blocks. 
From the second to the n+1'th lines , each line describing the i‐1'th block's a,b,c,d (1 =< ai,bi,ci <= 10^8 , d = 0 or 1 or 2). 
The input end with n = 0.

 

Output

Output a line contains a integer describing the highest "Skyscraper"'s height using the n blocks.

 

Sample Input

  

   3
10 10 12 0
10 10 12 1
10 10 11 2
2
10 10 11 1
10 10 11 1
0
  

 

Sample Output

  

   24
11
  

 

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

 

题意：给n个积木 每个积木有四个属性a b c d，a,b表示长与宽，长宽可以对调，c表示高度

d == 0 表示在垒积木的过程中 长与宽必须大于等于下面的积木的长与宽

d == 1 表示在垒积木的过程中 长与宽必须大于等于下面的积木的长于宽 并且面积必须大于下面的积木的面积

d == 2 表示在垒积木的过程中 长于宽必须大于下面的积木的长于宽

题目要求求出最高能垒出的高度

解题思路：可以设dp数组表示第i个积木为顶的最大高度，先对积木长宽排个序可以保证不会出现前面的积木可以垒在后面的积木上的情况，从第一个积木往后遍历 每次看看能不能给之前垒出的高度增高

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <climits>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

struct node
{
    int a,b,c,d;
    friend bool operator < (const node x,const node y)
    {
        if(x.a!=y.a) return x.a<y.a;
        if(x.b!=y.b) return x.b<y.b;
        return x.d>y.d;
    }
};
node x[1001];
LL dp[1001];

int main()
{
    int n;
    while(~scanf("%d",&n) && n)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d %d %d %d",&x[i].a,&x[i].b,&x[i].c,&x[i].d);
            if(x[i].a>x[i].b) swap(x[i].a,x[i].b);
        }
        sort(x,x+n);
        for(int i=0;i<n;i++)
        {
            dp[i]=x[i].c;
            for(int j=0;j<i;j++)
            {
                if(x[i].d==0)
                {
                    if(x[i].a>=x[j].a&&x[i].b>=x[j].b)
                        dp[i]=max(dp[i],dp[j]+x[i].c);
                }
                else if(x[i].d==1)
                {
                    if((x[i].a>x[j].a&&x[i].b>=x[j].b)||(x[i].a>=x[j].a&&x[i].b>x[j].b))
                        dp[i] = max(dp[i],dp[j]+x[i].c);
                }
                else if(x[i].a>x[j].a&&x[i].b>x[j].b)
                        dp[i]=max(dp[i],dp[j]+x[i].c);
            }
        }
        LL ans=0;
        for(int i=0;i<n;i++)
            ans = max(ans,dp[i]);
        printf("%lld\n",ans);
    }
    return 0;
}