HDU 1069 Monkey and Banana

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这个题好像以前在UVA上做过。反正就是以前见过这个题。

也是比较简单的dp了。对于每一种箱子都有三种方法。也就是说每一个边都可以当高。

并且每一种只会使用一次。所以把一个箱子所能旋转的所有情况都写出来。存起来。然后按照 x由大到小 y由大到小的顺序排序。

排序完之后 f【i】 = max（f【i】，f【j】 + v【i】）

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <algorithm>#include <fstream>#include <set>#include <map>#include <queue>#include <stack>#include <list>#include <vector>#include <cmath>#include <iomanip>#include <cmath>typedef long long LL;typedef unsigned long long LLU;const double PI=acos(-1.0);using namespace std;#define MAXN 1000+10#define INF 1 << 30struct Block{    int x,y;    int h;    Block(int a, int b, int c){x = a, y = b, h = c;}    friend bool operator < (Block a, Block b){        if(a.x != b.x)            return a.x > b.x;        else            return a.y > b.y;    }};int main (){    int n;    int kase = 0;    while(scanf("%d",&n) && n){        vector <Block> q;        for(int i = 0; i < n; i++){            int a[10];            scanf("%d%d%d",&a[0],&a[1],&a[2]);            sort(a,a+3);            if(a[0] == a[1] && a[1] == a[2])                q.push_back(Block(a[0],a[1],a[2]));            if(a[0] == a[1] && a[1] != a[2]){                q.push_back(Block(a[0],a[1],a[2]));                q.push_back(Block(a[0],a[2],a[1]));                q.push_back(Block(a[2],a[1],a[0]));            }            if(a[0] != a[1] && a[1] != a[2]){                q.push_back(Block(a[0],a[1],a[2]));                q.push_back(Block(a[0],a[2],a[1]));                q.push_back(Block(a[1],a[0],a[2]));                q.push_back(Block(a[1],a[2],a[0]));                q.push_back(Block(a[2],a[0],a[1]));                q.push_back(Block(a[2],a[1],a[0]));            }        }        sort(q.begin(), q.end());        LL f[MAXN] = {0};        int len = q.size();        for(int i = 0; i < len; i++)            f[i] = q[i].h;        for(int i = 0; i < len; i++){            for(int j = 0; j < i; j++){                if(q[i].x < q[j].x && q[i].y < q[j].y)                    f[i] = max(f[i], f[j]+q[i].h);            }        }        LL M = 0;        for(int i = 0; i < len; i++)            M = max(M, f[i]);        printf("Case %d: maximum height = %I64d\n",++kase,M);    }    return 0;}

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