HDU4001To Miss Our Children Time【dp】

最新推荐文章于 2020-03-06 20:07:24 发布

原创最新推荐文章于 2020-03-06 20:07:24 发布 · 293 阅读

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本文介绍了一种使用动态规划解决长方体木块堆叠问题的方法，通过排序和状态转移实现求解最大高度的目标。

大意：有三种长方体木块，只有满足如下关系一个木块才可以放在另一个的上面

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.

问最大高度

分析因为l和w都是一次递增的所以只要把其排好序那么就可以dp了后面的只会被前面的状态所决定

代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 const long long maxn = 1005;
 8 
 9 struct Node {
10     long long l, w, h, d;
11 }node[maxn];
12 long long dp[maxn];
13 bool cmp(Node n1, Node n2) {
14     if(n1.l != n2.l) return n1.l < n2.l;
15     else if(n1.w != n2.w) return n1.w < n2.w;
16     return n1.d > n2.d;
17 }
18 
19 int main() {
20     long long n;
21     while(scanf("%I64d",&n) ) {
22         if(n == 0) break;
23         memset(dp, 0, sizeof(dp));
24         for(long long i = 1; i <= n; i++) {
25             scanf("%I64d %I64d %I64d %I64d",&node[i].l, &node[i].w, &node[i].h, &node[i].d);
26             if(node[i].l < node[i].w) swap(node[i].l, node[i].w);
27         }
28         sort(node + 1, node + 1 + n, cmp);
29         for(long long i = 1; i <= n; i++) {
30             dp[i] = node[i].h;
31         }
32         for(long long i = 1; i <= n; i++) {
33             for(long long j = 1; j < i; j++) {
34                 if(node[i].d == 0) {
35                     if(node[i].l >= node[j].l && node[i].w >= node[j].w) {
36                         dp[i] = max(dp[i], dp[j] + node[i].h);
37                     } 
38                 } else if(node[i].d == 1) {
39                     if((node[i].l >= node[j].l && node[i].w > node[j].w) || (node[i].l > node[j].l && node[i].w >= node[j].w)) {
40                         dp[i] = max(dp[i], dp[j] + node[i].h);
41                     }
42                 } else {
43                     if(node[i].l > node[j].l && node[i].w > node[j].w) {
44                         dp[i] = max(dp[i], dp[j] + node[i].h);
45                     }
46                 }
47             }
48         }
49         long long ans = 0;
50         for(long long i = 1; i <= n; i++) ans = max(ans, dp[i]);
51         printf("%I64d\n", ans);
52     }
53     return 0;
54 }

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