HDU (1069)--monkey and Banana (LIS variant)

Test instructions

Now give you n a stone, then it is represented by coordinates (x, y, z). But it can have different methods, that is to say, its three coordinates can be rotated.

There is no limit to the number of stones, but each time it is necessary to keep the length and width of the bottom strictly decreasing, and then ask you what the maximum height you can spell with these stones.

Ideas:

Because there are a lot of coordinates, we can save every situation.

It is important to note that, at the time of saving, we have to keep X's coordinates greater than Y, so that we can sort them later.

Then it is just right to ask for the longest descending subsequence.

#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include < set> #include <map> #include <math.h>using namespace std; #define MAXN 111#define inf 99999999int DP[MAXN]; struct Node{int x,y,z,s;} A[maxn];bool CMP (node A,node b) {if (a.x!=b.x) return A.x>b.x;else if (a.x==b.x) return a.y>b.y;} int main () {int N;int j=1;while (~scanf ("%d", &n)) {if (n==0) break;int t=0;for (int i=0;i<n;i++) {int x,y,z;scanf ("% d%d%d ", &x,&y,&z);a[t].x=x>y?x:y; a[t].y=x>y?y:x;a[t].z=z; t++;a[t].x=y>z?y:z; a[t].y=y>z?z:y; A[t].z=x; t++;a[t].x=x>z?x:z; a[t].y=x>z?z:x; a[t].z=y;t++;} Sort (a,a+t,cmp); memset (Dp,0,sizeof (DP));DP [0]=a[0].z;int ans=-1;for (int i=0;i<t;i++) {int res=0;for (int j=0;j <i;j++) {if (a[i].x<a[j].x&&a[i].y<a[j].y) | | | (a[i].x<a[j].y&&a[i].y<a[j].x)) {Res=max (res,dp[j]);}} Dp[i]=res+a[i].z;ans=max (Ans,dp[i]);} printf ("Case%d:maximum height =%d\n", J++,ans);}}

Although the topic is simple, but this problem is also from my own thought out of, from easy to difficult to practice down!

I believe that my own efforts, in the near future will certainly be able to taste the joy of AC puzzles!

@ All the people who study DP, come on!!!

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HDU (1069)--monkey and Banana (LIS variant)