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Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13003????Accepted Submission(s): 6843

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.

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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.

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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".

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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 5 8 97 93 23 84 62 64 33 83 27 0

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Sample Output Case 1: maximum height = 40 Case 2: maximum height = 21 Case 3: maximum height = 28 Case 4: maximum height = 342

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題目鏈接：HDU 1069

題意：給出n種塊的尺寸，用坐標x、y、z表示，當某一個塊的兩個比下面的塊中某個兩個坐標小的時候就可以把這個塊疊上去，可以把塊旋轉一下使得三個坐標互相轉變……

塊的數量是無限的其實是嚇唬人的，由于是要嚴格上升，顯然三個已經把所有情況考慮了，四個肯定會跟某一種重復，然后每一個塊都假設有三個，用O（n²）的方法求一種可以說是帶權LIS的東西就行了

代碼：

#include <stdio.h> #include <bits/stdc++.h> using namespace std; #define INF 0x3f3f3f3f #define CLR(arr,val) memset(arr,val,sizeof(arr)) #define LC(x) (x<<1) #define RC(x) ((x<<1)+1) #define MID(x,y) ((x+y)>>1) typedef pair<int,int> pii; typedef long long LL; const double PI=acos(-1.0); const int N=190; struct info {int x;int y;int z;bool operator<(const info &t)const{if(x!=t.x)return x<t.x;return y<t.y;}info(int xx,int yy,int zz):x(xx),y(yy),z(zz){}info(){} }; vector<info>arr; int dp[N]; int main(void) {int n,i,j,x,y,z;int q=1;while (~scanf("%d",&n)&&n){CLR(dp,0);arr.clear();for (i=0; i<n; ++i){scanf("%d%d%d",&x,&y,&z);arr.push_back(info(x,y,z));arr.push_back(info(x,z,y));arr.push_back(info(y,x,z));arr.push_back(info(y,z,x));arr.push_back(info(z,x,y));arr.push_back(info(z,y,x));}sort(arr.begin(),arr.end());int SZ=arr.size();for (i=0; i<SZ; ++i){int pre_max=0;for (j=0; j<i; ++j){if(arr[j].x<arr[i].x&&arr[j].y<arr[i].y&&dp[j]>pre_max)pre_max=dp[j];}dp[i]=pre_max+arr[i].z;}printf("Case %d: maximum height = %d\n",q++,*max_element(dp,dp+SZ));//n^2算法這里就要取max，之前搞混了直接輸出最后一個導致WA}return 0; }

轉載于:https://www.cnblogs.com/Blackops/p/5930755.html

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