一个盒子的高有三种状态,只要比较长和宽就行了。n个正方体可以看成n×3个矩形嵌套问题。可以转化成有向无环图,建完图后用记忆化搜索即可。
状态转移方程:d(i) = max{d(j) + box[i].z}
代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 30 * 3 + 10;
struct Box {
int x;
int y;
int z;
} box[maxn];
int d[maxn];
bool G[maxn][maxn];
int n;
int dp(int i) {
int& ans = d[i];
if (ans > 0) return ans;
ans = box[i].z;
for(int j = 1; j <= 3*n; j++) {
if (G[i][j]) ans = max(ans, dp(j) + box[i].z);
}
return ans;
}
bool judge(Box a, Box b) { // 判断a能否连到b
return (a.x > b.x && a.y > b.y || a.x > b.y && a.y > b.x);
}
int main() {
int kase = 0;
freopen("input.txt", "r", stdin);
while(~scanf("%d", &n) && n) {
memset(G, false, sizeof(G));
memset(d, -1, sizeof(d));
for(int i = 1; i < 3 * n; i += 3) {
scanf("%d%d%d", &box[i].x, &box[i].y, &box[i].z);
box[i+1].x = box[i].z;
box[i+1].y = box[i].x;
box[i+1].z = box[i].y;
box[i+2].x = box[i].y;
box[i+2].y = box[i].z;
box[i+2].z = box[i].x;
}
for(int i = 1; i <= 3 * n; i++) {
for(int j = 1; j <= 3 * n; j++) {
if (i == j) continue;
else {
if (judge(box[i], box[j])) {
G[i][j] = true;
}
}
}
}
int ans = -0x7fffffff;
for(int i = 1; i <= 3 * n; i++) {
ans = max(ans, dp(i));
}
printf("Case %d: maximum height = %d\n", ++kase, ans);
}
return 0;
}