Uva 437-The Tower of Babylon(DP)

2015-01-24 05:42:41 · 作者: · 浏览: 3

题目链接:点击打开链接

DAG上最长路、

题意:给出n种长方体,(每种无限),要求能摞的最大高度。连边是大的连小的。

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       #include 
      
        #include 
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              #include 
              #include 
              
                #define maxn 10000002 #define _ll __int64 #define ll long long #define INF 0x3f3f3f3f #define Mod 10000007 #define pp pair
               
                 #define ull unsigned long long using namespace std; int n,m,dp[110],x[110],y[110],z[110]; bool ma[110][110]; void input() { m=n*3; for(int i=1;i<=m-2;i+=3){ scanf("%d %d %d",x+i,y+i,z+i); x[i+1]=y[i];y[i+1]=z[i];z[i+1]=x[i]; x[i+2]=y[i+1];y[i+2]=z[i+1];z[i+2]=x[i+1]; } } void build() { memset(ma,0,sizeof(ma)); memset(dp,0,sizeof(dp)); for(int i=1;i<=m;i++) for(int j=i+1;j<=m;j++){ if((x[i]>x[j]&&y[i]>y[j])||(x[i]>y[j]&&y[i]>x[j])) ma[i][j]=1; else if((x[j]>x[i]&&y[j]>y[i])||(x[j]>y[i]&&y[j]>x[i])) ma[j][i]=1; } } int dfs(int x) { int& ans=dp[x]; if(ans)return ans; ans=z[x]; for(int i=1;i<=m;i++){ if(ma[x][i])ans=max(ans,dfs(i)+z[x]); } return ans; } int main() { int cas=1; while(scanf("%d",&n)&&n){ input(); build(); int ans=-INF; for(int i=1;i<=m;i++) ans=max(ans,dfs(i)); printf("Case %d: maximum height = %d\n",cas++,ans); } return 0; }