Monkey and Banana
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5406 Accepted Submission(s): 2769
Problem Description
A group of researchers are designing an experiment to test the IQ of a monkey. They will hang a banana at the roof of a building, and at the mean time, provide the monkey with some blocks. If the monkey is clever enough, it shall be able to reach the banana by placing one block on the top another to build a tower and climb up to get its favorite food.
The researchers have n types of blocks, and an unlimited supply of blocks of each type. Each type-i block was a rectangular solid with linear dimensions (xi, yi, zi). A block could be reoriented so that any two of its three dimensions determined the dimensions of the base and the other dimension was the height.
They want to make sure that the tallest tower possible by stacking blocks can reach the roof. The problem is that, in building a tower, one block could only be placed on top of another block as long as the two base dimensions of the upper block were both strictly smaller than the corresponding base dimensions of the lower block because there has to be some space for the monkey to step on. This meant, for example, that blocks oriented to have equal-sized bases couldn't be stacked.
Your job is to write a program that determines the height of the tallest tower the monkey can build with a given set of blocks.
The researchers have n types of blocks, and an unlimited supply of blocks of each type. Each type-i block was a rectangular solid with linear dimensions (xi, yi, zi). A block could be reoriented so that any two of its three dimensions determined the dimensions of the base and the other dimension was the height.
They want to make sure that the tallest tower possible by stacking blocks can reach the roof. The problem is that, in building a tower, one block could only be placed on top of another block as long as the two base dimensions of the upper block were both strictly smaller than the corresponding base dimensions of the lower block because there has to be some space for the monkey to step on. This meant, for example, that blocks oriented to have equal-sized bases couldn't be stacked.
Your job is to write a program that determines the height of the tallest tower the monkey can build with a given set of blocks.
Input
The input file will contain one or more test cases. The first line of each test case contains an integer n,
representing the number of different blocks in the following data set. The maximum value for n is 30.
Each of the next n lines contains three integers representing the values xi, yi and zi.
Input is terminated by a value of zero (0) for n.
representing the number of different blocks in the following data set. The maximum value for n is 30.
Each of the next n lines contains three integers representing the values xi, yi and zi.
Input is terminated by a value of zero (0) for n.
Output
For each test case, print one line containing the case number (they are numbered sequentially starting from 1) and the height of the tallest possible tower in the format "Case case: maximum height = height".
Sample Input
1
10 20 30
2
6 8 10
5 5 5
7
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
5
31 41 59
26 53 58
97 93 23
84 62 64
33 83 27
0
Sample Output
Case 1: maximum height = 40
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342
1 #include<stdio.h> 2 #include<algorithm> 3 const int maxn = 200; 4 struct Node 5 { 6 int x,y,z; 7 int dp; 8 }stick[maxn]; 9 int cmp(const struct Node a, const struct Node b) 10 { 11 if(a.x != b.x) 12 return a.x < b.x; 13 else if(a.x == b.x) return a.y < b.y; 14 else return 0; 15 16 } 17 int main() 18 { 19 int n,x,y,z,k; 20 int cnt = 1; 21 while(~scanf("%d",&n)&&n) 22 { 23 k = 0; 24 for(int i = 0; i < n; i++) 25 { 26 scanf("%d %d %d",&x,&y,&z); 27 if(x == y) 28 { 29 if(y == z) 30 { 31 stick[k].x = x; 32 stick[k].y = y; 33 stick[k].z = z; 34 stick[k].dp = stick[k].z; 35 k++; 36 } 37 else 38 { 39 stick[k].x = x; 40 stick[k].y = y; 41 stick[k].z = z; 42 stick[k].dp = stick[k].z; 43 k++; 44 stick[k].x = z; 45 stick[k].y = y; 46 stick[k].z = x; 47 stick[k].dp = stick[k].z; 48 k++; 49 stick[k].x = y; 50 stick[k].y = z; 51 stick[k].z = x; 52 stick[k].dp = stick[k].z; 53 k++; 54 55 56 } 57 } 58 else 59 { 60 if(y == z) 61 { 62 stick[k].x = x; 63 stick[k].y = y; 64 stick[k].z = z; 65 stick[k].dp = stick[k].z; 66 k++; 67 stick[k].x = z; 68 stick[k].y = y; 69 stick[k].z = x; 70 stick[k].dp = stick[k].z; 71 k++; 72 stick[k].x = y; 73 stick[k].y = x; 74 stick[k].z = z; 75 stick[k].dp = stick[k].z; 76 k++; 77 } 78 else if(x == z) 79 { 80 stick[k].x = x; 81 stick[k].y = y; 82 stick[k].z = z; 83 stick[k].dp = stick[k].z; 84 k++; 85 stick[k].x = y; 86 stick[k].y = z; 87 stick[k].z = x; 88 stick[k].dp = stick[k].z; 89 k++; 90 stick[k].x = x; 91 stick[k].y = z; 92 stick[k].z = y; 93 stick[k].dp = stick[k].z; 94 k++; 95 } 96 else 97 { 98 stick[k].x = x; 99 stick[k].y = y; 100 stick[k].z = z; 101 stick[k].dp = stick[k].z; 102 k++; 103 stick[k].x = x; 104 stick[k].y = z; 105 stick[k].z = y; 106 stick[k].dp = stick[k].z; 107 k++; 108 stick[k].x = y; 109 stick[k].y = x; 110 stick[k].z = z; 111 stick[k].dp = stick[k].z; 112 k++; 113 stick[k].x = y; 114 stick[k].y = z; 115 stick[k].z = x; 116 stick[k].dp = stick[k].z; 117 k++; 118 stick[k].x = z; 119 stick[k].y = x; 120 stick[k].z = y; 121 stick[k].dp = stick[k].z; 122 k++; 123 stick[k].x = z; 124 stick[k].y = y; 125 stick[k].z = x; 126 stick[k].dp = stick[k].z; 127 k++; 128 } 129 } 130 131 } 132 std::sort(stick,stick+k,cmp); 133 int maxh = 0,i,j; 134 for( i = 1;i < k; i++) 135 { 136 for(j = 0; j < i; j++) 137 { 138 if(stick[i].x > stick[j].x&&stick[i].y>stick[j].y&& 139 stick[i].dp < stick[j].dp+stick[i].z) 140 stick[i].dp = stick[j].dp + stick[i].z; 141 } 142 if(maxh < stick[i].dp) 143 maxh = stick[i].dp; 144 } 145 printf("Case %d: maximum height = %d\n",cnt++,maxh); 146 } 147 return 0; 148 }
题目：给出一些长方体，然后让你把他堆成塔，
要求下面的塔的要比上面的塔大（长和宽），
而且每一种长方体的数量都是无限的。
此题目考察到动态规划里的最长有序子序列
每输入三个数x,y,z,把他们所可能对应长宽高的情况放在数组里，这样解决了每个长方体都无限多的问题，然后转化为求最长上升子序列的问题。是一道经典dp的问题。