当看懂这个题意的时候，就没啥难度了，百练上有个中文版的数据一模一样。。。普通O（n^2)DP。

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 int o[2000]; 5 struct node 6 { 7     int x,y; 8     int h; 9 }p[2000];10 int cmp(const void *a,const void *b)11 {12     return (*(struct node *)a).x >(*(struct node *)b).x ? 1:-1;13 }14 int main()15 {16     int n,i,j,num = 0,sv,ev,w,max;17     while(scanf("%d",&n)!=EOF)18     {19         if(!n) break;20         num ++;21         memset(p,0,sizeof(p));22         memset(o,0,sizeof(o));23         for(i = 1;i <= n;i ++)24         {25             scanf("%d%d%d",&sv,&ev,&w);26             p[6*(i-1)].x = sv;27             p[6*(i-1)].y = ev;28             p[6*(i-1)].h = w;29             p[6*(i-1)+1].x = ev;30             p[6*(i-1)+1].y = sv;31             p[6*(i-1)+1].h = w;32             p[6*(i-1)+2].x = ev;33             p[6*(i-1)+2].y = w;34             p[6*(i-1)+2].h = sv;35             p[6*(i-1)+3].x = w;36             p[6*(i-1)+3].y = sv;37             p[6*(i-1)+3].h = ev;38             p[6*(i-1)+4].x = w;39             p[6*(i-1)+4].y = ev;40             p[6*(i-1)+4].h = sv;41             p[6*(i-1)+5].x = sv;42             p[6*(i-1)+5].y = w;43             p[6*(i-1)+5].h = ev;44         }45         qsort(p,6*n,sizeof(p[0]),cmp);46         o[0] = p[0].h;47         for(i = 1;i <= n*6-1;i ++)48         {49             max = 0;50             for(j = 0;j <= i-1;j ++)51             {52                 if(p[i].x > p[j].x&&p[i].y > p[j].y)53                 {54                     if(max < o[j])55                     max = o[j];56                 }57                 o[i] = max+p[i].h;58             }59         }60         max = 0;61         for(i = 0;i <= 6*n-1;i ++)62         {63             if(max < o[i])64             max = o[i];65         }66         printf("Case %d: maximum height = %d\n",num,max);67     }68     return 0;69 }