HDU - 1069 Monkey and Banana(DAG)

题目大意：给你N种积木，每种的数量不限，现在要求你求出能用这些积木堆出的最大高度
堆的规则如下：在底下的积木的的长和宽要大于在其上面的所有积木的长和宽

解题思路：一种积木有6种摆放方式，所以先预处理出每种积木的所有摆放方式
因为长宽限制，所以每种积木的每种摆放方式最多只有一块
接着排个序，按照面积排序，接下来就是纯DAG问题了

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 210;struct Block{    int x, y, z;}B[N];int dp[N];int n;bool cmp(const Block &a, const Block &b) {    return a.x * a.y > b.x * b.y;}void init() {    //1 2 3    for (int i = 0; i < n; i++)         scanf("%d%d%d", &B[i].x, &B[i].y, &B[i].z);    int cnt = n;    for (int i = 0; i < n; i++) {        B[cnt].x = B[i].x; B[cnt].y = B[i].z; B[cnt++].z = B[i].y;  //1 3 2        B[cnt].x = B[i].y; B[cnt].y = B[i].x; B[cnt++].z = B[i].z;  //2 1 3        B[cnt].x = B[i].y; B[cnt].y = B[i].z; B[cnt++].z = B[i].x;  //2 3 1        B[cnt].x = B[i].z; B[cnt].y = B[i].x; B[cnt++].z = B[i].y;  //3 1 2         B[cnt].x = B[i].z; B[cnt].y = B[i].y; B[cnt++].z = B[i].x;  //3 2 1    }    n = cnt;}int cas = 1;void solve() {    sort(B, B + n, cmp);    for (int i = 0; i < n; i++)        dp[i] = B[i].z;    for (int i = 0; i < n; i++)        for (int j = 0; j < i; j++)             if (B[i].x < B[j].x && B[i].y < B[j].y)                 dp[i] = max(dp[i], dp[j] + B[i].z);    int ans = 0;    for (int i = 0; i < n; i++)        ans = max(ans, dp[i]);    printf("Case %d: maximum height = %d\n", cas++, ans);}int main() {    while (scanf("%d", &n) != EOF && n) {        init();        solve();    }    return 0;}