【HDOJ】1069 Monkey and Banana

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 <stdio.h>
#include <stdlib.h>

#define MAXTYPE 33
#define MAXNUM  3*MAXTYPE

typedef 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;

    //
    tmp = (y<z) ? y:z;
    rects[i+].x = tmp;
    rects[i+].y = y+z-tmp;
    rects[i+].z = x;

    //
    tmp = (x<z) ? x:z;
    rects[i+].x = tmp;
    rects[i+].y = x+z-tmp;
    rects[i+].z = y;

    //
    tmp = (x<y) ? x:y;
    rects[i].x = tmp;
    rects[i].y = x+y-tmp;
}

int mymax(int a, int b) {
    return a>b ? a:b;
}

int main() {
    int n, total;
    int i, j, tmp;
    int case_n = ;

    while (scanf("%d", &n)!=EOF && n) {
        case_n ++;

        for (i=; i<*n; i=i+) {
            scanf("%d %d %d", &rects[i].x, &rects[i].y, &rects[i].z);
            addOth(i);
        }

        total = *n;
        qsort(rects, total, sizeof(rect), comp);

        /*
        for (i=0; i<total; ++i)
            printf("%d: %d, %d, %d\n", i, rects[i].x, rects[i].y, rects[i].z);
        */

        heights[] = rects[].z;
        for (i=; i<total; ++i) {
            tmp = ;
            for (j=i-; j>=; --j) {
                if (rects[j].x<rects[i].x && rects[j].y<rects[i].y && heights[j]>tmp)
                    tmp = heights[j];
            }
            heights[i] = mymax(tmp+rects[i].z, rects[i].z);
        }

        tmp = ;
        for (i=; i<total; ++i) {
            if (heights[i] > tmp)
                tmp = heights[i];
        }

        printf("Case %d: maximum height = %d\n", case_n, tmp);
    }

    return ;
}