[2016-03-30][HDU][1069][Monkey and Banana]

• 时间:2016-03-27 15:19:40 星期日

• 题目编号:[2016-03-30][HDU][1069][Monkey and Banana]

• 题目大意:给定n种积木无限个,问这些积木最大能叠多高,上面的积木长宽必须严格小于下面的积木

• 分析:

  • dp[i]表示第i个积木在顶部时候的最大高度,那么dp[i] = max(dp[i],dp[j] + h[i]);∀ji能放在j上面$\mathrm{\forall }ji能放在j上面$
  • 初始条件就是长宽最大的高度是它自己,

#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long LL;
int dp[90*90];
struct mRect{
        int x,y,z;
        mRect(int a = 0,int b = 0,int c = 0):x(a),y(b),z(c){};
        bool operator < (const mRect & a)const{
                return x > a.x || (x == a.x && y > a.y) || (x == a.x && y == a.y && z > a.z);
        }
        bool operator == (const mRect & a)const{
                return x == a.x && y == a.y && z == a.z;
        }
}r[90 * 90];
int main(){
        int n,cntcase = 0;
        while(~scanf("%d",&n) && n){
                int a[3],cur = 1;
                for(int i = 0 ;i < n ; ++i){
                        for(int j = 0;j < 3 ;++j)       scanf("%d",&a[j]);
                        sort(a,a+3);
                        r[cur++] = mRect(a[0],a[1],a[2]);
                        r[cur++] = mRect(a[0],a[2],a[1]);
                        r[cur++] = mRect(a[1],a[2],a[0]);
                }
                sort(r+1,r+cur);
                cur = unique(r+1,r+cur) - r;
                memset(dp,0,sizeof(dp));
                r[0] = mRect(0x3f3f3f3f,0x3f3f3f3f,0);
                int maxh = 0;
                for(int i = 1;i < cur;++i){
                        for(int j = 0;j < i;++j){
                                if(r[i].x < r[j].x && r[i].y < r[j].y){
                                        dp[i] = max(dp[i],dp[j] + r[i].z);
                                }
                        }
                        maxh = max(maxh,dp[i]);
                }
                printf("Case %d: maximum height = %d\n",++cntcase,maxh);
        }
        return 0;
}

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