【UVA 437】The Tower of Babylon(拓扑排序+DP,做法)

【Solution】

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;

f[y]=max(f[y],f[x]+h[y]);(x−>y)∈E

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max(f[1..n])

【Code】

#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\n",d);
    }
    return 0;
}