HDOJ 1069 Monkey and Banana

题意：在高度为h处有一串香蕉，猴子想要拿到它，但是他的高度不够，需要通过长方体木块的堆积才能拿到，已知有n钟长宽高分别为x,y,z的长方体木块，每种木块的个数足够多，判断这些木块可以堆叠的最大高度

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1069

思路：动态规划。对每个长方形进行六种旋转，在对底上的中一条边进行排序，求出最大可能。

注意点：旋转方向六种，分别为(x,y,z),(x,z,y),(y,x,z),(y,z,x),(z,x,y),(z,y,x)

以下为AC代码：

Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor116225902014-09-09 00:53:51Accepted106915MS412K2080 BG++luminous11

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <iostream>#include <cstring>#include <cstdio>#include <string>#include <vector>#include <deque>#include <list>#include <map>#include <set>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <cctype>#include <climits>#include <iomanip>#include <cstdlib>#include <algorithm>//#include <unordered_map>//#include <unordered_set>#define ll long long#define ull unsigned long long#define all(x) (x).begin(), (x).end()#define clr(a, v) memset( a , v , sizeof(a) )#define pb push_back#define RDI(a) scanf ( "%d", &a )#define RDII(a, b) scanf ( "%d%d", &a, &b )#define RDIII(a, b, c) scanf ( "%d%d%d", &a, &b, &c )#define RS(s) scanf ( "%s", s )#define PI(a) printf ( "%d", a )#define PIL(a) printf ( "%d\n", a )#define PII(a,b) printf ( "%d %d", a, b )#define PIIL(a,b) printf ( "%d %d\n", a, b )#define PIII(a,b,c) printf ( "%d %d %d", a, b, c )#define PIIIL(a,b,c) printf ( "%d %d %d\n", a, b, c )#define PL() printf ( "\n" )#define PSL(s) printf ( "%s\n", s )#define rep(i,m,n) for ( int i = m; i <  n; i ++ )#define REP(i,m,n) for ( int i = m; i <= n; i ++ )#define dep(i,m,n) for ( int i = m; i >  n; i -- )#define DEP(i,m,n) for ( int i = m; i >= n; i -- )#define repi(i,m,n,k) for ( int i = m; i <  n; i += k )#define REPI(i,m,n,k) for ( int i = m; i <= n; i += k )#define depi(i,m,n,k) for ( int i = m; i >  n; i += k )#define DEPI(i,m,n,k) for ( int i = m; i >= n; i -= k )#define READ(f) freopen(f, "r", stdin)#define WRITE(f) freopen(f, "w", stdout)using namespace std;const double pi = acos(-1);template <class T>inline bool RD ( T &ret ){    char c;    int sgn;    if ( c = getchar(), c ==EOF )return 0; //EOF    while ( c != '-' && ( c < '0' || c > '9' ) ) c = getchar();    sgn = ( c == '-' ) ? -1 : 1;    ret = ( c == '-' ) ? 0 : ( c - '0' );    while ( c = getchar() , c >= '0' && c <= '9' ) ret = ret * 10 + ( c - '0' );    ret *= sgn;    return 1;}inline void PD ( int x ){    if ( x > 9 ) PD ( x / 10 );    putchar ( x % 10 + '0' );}const double eps = 1e-10;const int dir[4][2] = { 1,0, -1,0, 0,1, 0,-1 };struct node{    int x, y, z, s, cnt;    node(){}    node( int _cnt ) : cnt(_cnt) {}    node( int _x, int _y, int _z ) : x(_x), y(_y), z(_z), cnt(_z), s(_x*_y) {}    node( int _x, int _y, int _z, int _cnt ) : x(_x), y(_y), z(_z), cnt(_cnt) {}    friend bool operator < ( const node &a, const node &b ){        if ( a.x != b.x )            return a.x > b.x;        else            return a.y > b.y;    }};vector<node> k;int main(){    int n;    int ncase = 1;    while ( RDI ( n ) != EOF && n ){        k.clear();        int x, y, z;        rep ( i, 0, n ){            cin >> x >> y >> z;            k.push_back( node ( x, y, z ) );            k.push_back( node ( y, x, z ) );            k.push_back( node ( y, z, x ) );            k.push_back( node ( z, y, x ) );            k.push_back( node ( x, z, y ) );            k.push_back( node ( z, x, y ) );        }        sort ( all ( k ) );        int ans = 0;        rep ( i, 0, k.size() ){            rep ( j, 0, i ){                if ( k[i].x < k[j].x && k[i].y < k[j].y ){                    k[i].cnt = max ( k[i].cnt, k[j].cnt + k[i].z );                    ans = max ( ans, k[i].cnt );                }            }        }        printf ( "Case %d: maximum height = %d\n", ncase ++, ans );    }    return 0;}