接上一篇,在处理有向无环图的最长链问题的时候,可以在做拓扑排序的同时,一边做DP;
设f[i]表示第i个方块作为最上面的最高值;
这样可以保证,按阶段进行DP,每次在获取f[x]的时候,你可以保证f[x]已经获得了;
最后取
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define Open() freopen("D:\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 30;
struct abc{
LL c,k,g;
};
int n,b[4],nn,du[N*3+100];
LL dp[N*3+100];
abc a[N*3+100];
vector <int> G[N*3+100];
queue <int> dl;
int main()
{
//Open();
int kk = 0;
while (~scanf("%d",&n) && n){
kk++;
ms(dp,-1);nn = 0;ms(du,0);
rep1(i,1,N*3) G[i].clear();
rep1(i,1,n){
rep1(j,1,3)
scanf("%d",&b[j]);
sort(b+1,b+1+3);
rep1(j,1,3){
nn++;
rep2(k,3,1)
if (k!=j){
a[nn].c = b[k];
break;
}
rep1(k,1,3)
if (k!=j){
a[nn].k = b[k];
break;
}
a[nn].g = b[j];
}
}
n = nn;
rep1(i,1,n)
rep1(j,1,n)
if (a[i].c > a[j].c && a[i].k > a[j].k){
G[i].pb(j);
du[j]++;
}
while (!dl.empty()) dl.pop();
rep1(i,1,n)
if (du[i]==0){
dl.push(i);
dp[i] = a[i].g;
du[i] = -1;
}
while (!dl.empty()){
int x = dl.front();
dl.pop();
int len = G[x].size();
rep1(i,0,len-1){
int y = G[x][i];
if (dp[y]==-1){
dp[y] = dp[x] + a[y].g;
}else
dp[y] = max(dp[y],dp[x]+a[y].g);
du[y]--;
if (du[y]==0){
dl.push(y);
du[y]= -1;
}
}
}
LL d = 0;
rep1(i,1,n)
d = max(d,dp[i]);
printf("Case %d: maximum height = ",kk);
printf("%lld
",d);
}
return 0;
}