To Miss Our Children Time

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 4075    Accepted Submission(s): 1063

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

卡了一下check，在宽>长那里 。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>
#include <vector>
#include <queue>

using namespace std ;
typedef long long LL ;
typedef pair<int,int> pii;
#define X first
#define Y second
const int N = 1010;
struct Blocks {
    LL a , b , c , d , area ;
    bool operator < ( const Blocks &A ) const {
        if( a != A.a ) return a < A.a ;
        else if( b != A.b ) return b < A.b ;
        else return d > A.d ;
    }
}e[N];
LL dp[N] ;
int main ()  {
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    int _ , cas =1 , n ;
    while( cin >> n && n ) {
        memset( dp , 0 , sizeof dp );
        for( int i = 0 ; i < n ; ++i ) {
            cin >> e[i].a >> e[i].b >> e[i].c >> e[i].d ;
            if(e[i].a > e[i].b ) swap(e[i].a, e[i].b );
            e[i].area =  e[i].a * e[i].b ;
        }
        sort( e  , e + n  ) ;
        for( int i = 0 ; i < n ; ++i ) {
            dp[i] = e[i].c ;
            for( int j = 0 ; j < i ; ++j ) {
                if( e[i].d == 0 && e[j].a <= e[i].a && e[j].b <= e[i].b )
                    dp[i] = max( dp[i] , dp[j] + e[i].c );
                if( e[i].d == 1 && e[j].a <= e[i].a && e[j].b <= e[i].b && e[j].area < e[i].area )
                    dp[i] = max( dp[i] , dp[j] + e[i].c );
                if( e[i].d == 2 && e[j].a < e[i].a && e[j].b < e[i].b )
                    dp[i] = max( dp[i] , dp[j] + e[i].c );
            }
        }
        LL ans = 0 ; for( int i = 0 ; i < n ; ++i ) ans = max( ans , dp[i] );
        cout << ans << endl ;
    }
}