HDU 1069 dp最长递增子序列
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Description
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
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
numbered sequentially starting from 1) and the height of the tallest
possible tower in the format "Case case: maximum height = height".
Sample Input
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 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct node{
int x,y,z;
}que[];
int dp[];
int tot; void addedge(int x,int y,int z){
que[tot].x=x;
que[tot].y=y;
que[tot++].z=z; que[tot].x=x;
que[tot].y=z;
que[tot++].z=y; que[tot].x=y;
que[tot].y=x;
que[tot++].z=z; que[tot].x=y;
que[tot].y=z;
que[tot++].z=x; que[tot].x=z;
que[tot].y=x;
que[tot++].z=y; que[tot].x=z;
que[tot].y=y;
que[tot++].z=x;
} bool cmp(struct node t1,struct node t2){
if(t1.x!=t2.x)
return t1.x>t2.x;
else if(t1.x==t2.x&&t1.y!=t2.y)
return t1.y>t2.y;
else
return t1.z>t2.z;
} int main(){
int n;
int cas=;
while(scanf("%d",&n)!=EOF){
if(!n)
break;
memset(dp,,sizeof(dp));
tot=;
int x,y,z;
for(int i=;i<n;i++){
scanf("%d%d%d",&x,&y,&z);
addedge(x,y,z);
}
sort(que+,que+tot+,cmp);
dp[]=que[].z;
for(int i=;i<tot;i++){
dp[i]=que[i].z;
for(int j=i-;j>=;j--){
if(que[i].x<que[j].x&&que[i].y<que[j].y&&dp[i]<dp[j]+que[i].z)
dp[i]=dp[j]+que[i].z;
}
}
int ans=-;
for(int i=;i<tot;i++)
ans=max(ans,dp[i]);
printf("Case %d: maximum height = %d\n",cas++,ans);
}
return ;
}
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