DP问题，我是按照边排序的，排序既要考虑x也要考虑y，同时在每个面中，长宽也要有序。还有注意状态转移，当前高度并不是之前的最大block叠加的高度，而是可叠加最大高度+当前block高度或者是当前block高度。最后，n^2的复杂度。

#include 
     
      #include 
      
       #define MAXTYPE 33#define MAXNUM  3*MAXTYPEtypedef struct {    int x, y, z;} rect;rect rects[MAXNUM];int heights[MAXNUM];int comp(const void *a, const void *b) {    if ( ((rect *)a)->x - ((rect *)b)->x )        return ((rect *)a)->x - ((rect *)b)->x;    else        return ((rect *)a)->y - ((rect *)b)->y;}void addOth(int i) {    int tmp;    int x = rects[i].x;    int y = rects[i].y;    int z = rects[i].z;    // 2    tmp = (y
       
        b ? a:b;}int main() {    int n, total;    int i, j, tmp;    int case_n = 0;    while (scanf("%d", &n)!=EOF && n) {        case_n ++;        for (i=0; i<3*n; i=i+3) {            scanf("%d %d %d", &rects[i].x, &rects[i].y, &rects[i].z);            addOth(i);        }        total = 3*n;        qsort(rects, total, sizeof(rect), comp);        /*        for (i=0; i
        
         =0; --j) {                if (rects[j].x
         
          tmp)                    tmp = heights[j];            }            heights[i] = mymax(tmp+rects[i].z, rects[i].z);        }        tmp = 0;        for (i=0; i
          
            tmp) tmp = heights[i]; } printf("Case %d: maximum height = %d\n", case_n, tmp); } return 0;}