HDU

题意：已知n种砖，每种砖有无穷多个，问这些砖最多能堆多高，要求堆放时，上面的砖摆放的长和宽必须都严格小于下面砖的长和宽。

分析：

1、将砖堆放时的堆放面上所有的长x和宽y统计出来，去重，且记录时保证x<y。每种砖最多能统计出三种长和宽。

2、如果之前的砖比当前砖长和宽都严格小，则可将之前的砖放在这个砖上面dp[i] = max(dp[i], dp[j] + num[i].h);

3、dp[i]---截止到第i块砖，能堆的最大高度。

4、因为不知道哪个砖放在最底下，所以扫一遍，ans = max(ans, dp[i]);

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
map<pair<int, int>, int> mp;
struct Node{
    int x, y, h;
}num[110];
int dp[110];
int main(){
    int N;
    int kase = 0;
    while(scanf("%d", &N) == 1){
        if(!N) return 0;
        memset(dp, 0, sizeof dp);
        mp.clear();
        int x, y, z;
        for(int i = 0; i < N; ++i){
            scanf("%d%d%d", &x, &y, &z);
            vector<int> v;
            v.push_back(x);
            v.push_back(y);
            v.push_back(z);
            sort(v.begin(), v.end());
            if(!mp.count(pair<int, int>(v[0], v[1]))){
                mp[pair<int, int>(v[0], v[1])] = v[2];
            }
            if(!mp.count(pair<int, int>(v[1], v[2]))){
                mp[pair<int, int>(v[1], v[2])] = v[0];
            }
            if(!mp.count(pair<int, int>(v[0], v[2]))){
                mp[pair<int, int>(v[0], v[2])] = v[1];
            }
        }
        int cnt = 0;
        for(map<pair<int, int>, int>::iterator it = mp.begin(); it != mp.end(); ++it){
            num[cnt].x = (*it).first.first;
            num[cnt].y = (*it).first.second;
            num[cnt].h = (*it).second;
            ++cnt;
        }
        for(int i = 0; i < cnt; ++i){
            dp[i] = num[i].h;
            for(int j = 0; j < i; ++j){
                if(num[j].x < num[i].x && num[j].y < num[i].y){
                    dp[i] = max(dp[i], dp[j] + num[i].h);
                }
            }
        }
        int ans = 0;
        for(int i = 0; i < cnt; ++i){
            ans = max(ans, dp[i]);
        }
        printf("Case %d: maximum height = %d
", ++kase, ans);
    }
    return 0;
}