To Miss Our Children Time(dp)

Posted 2020-07-11 handsomecui

To Miss Our Children Time

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 4502    Accepted Submission(s): 1224

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

题解：当时没细想，就是个很水的dp。。。1000两重for，注意long long；

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
const int MAXN = 1010;
typedef long long LL;
struct Node{
    int w, h, c, d;
    void init(){
        scanf("%d%d%d%d", &w, &h, &c, &d);
        if(w > h){
            swap(w, h);
        }
    }
    friend bool operator < (Node a, Node b){
        if(a.h != b.h)
            return a.h < b.h;
        if(a.w !=  b.w)
            return a.w < b.w;
        return a.d > b.d;
    }
};
Node dt[MAXN];
LL dp[MAXN];
bool js(Node a, Node b){
    if(b.d == 2 && a.w < b.w && a.h < b.h)
        return true;
    else if(b.d == 1 && a.w <= b.w && a.h <= b.h && (a.w < b.w || a.h < b.h))
        return true;
    else if(b.d == 0 && a.w <= b.w && a.h <= b.h)
        return true;
    return false;
}
int main(){
    int n;
    while(~scanf("%d", &n), n){
        for(int i = 0; i < n; i++){
            dt[i].init();
        }
        sort(dt, dt + n);
        memset(dp, 0, sizeof(dp));
        LL ans = 0;
        for(int i = 0; i < n; i++){
              dp[i] = dt[i].c;
            for(int j = 0; j < i; j++){
                if(js(dt[j], dt[i]))
                    dp[i] = max(dp[i], dp[j] + dt[i].c);
            }
            ans = max(ans, dp[i]);
        }
        printf("%lld\n", ans);
    }
    return 0;
}