HDU 1069 Monkey and Banana(DP）
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 <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 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 }
