UVa 437. The Tower of Babylon

把所有的正方体形态全求一遍，然后就是裸DAG了

#include <iostream>#include <algorithm>#include <cstdio>#include <cstdlib>#include <cstring>#include <set>using namespace std;const int INF = 0x7fffffff;const int maxn = 300;struct node{    int c, k, g;    void sc(int x, int y, int z) {        c = x, k = y, g = z;    }    int S() { return c * k; }    bool operator<(const node& a) const {        return c * k < a.c * a.k;    }}str[maxn];int n, dp[maxn], kase = 0;int main(){    while(scanf("%d", &n) != EOF && n) {        int c, k, g, cnt = 0;        memset(dp, 0, sizeof(dp));        for(int i = 0; i < n; ++i) {            scanf("%d%d%d", &c, &k, &g);            str[cnt++].sc(c, k, g);            str[cnt++].sc(c, g, k);            str[cnt++].sc(g, c, k);            str[cnt++].sc(g, k, c);            str[cnt++].sc(k, g, c);            str[cnt++].sc(k, c, g);        }        sort(str, str+cnt);        for(int i = 0; i < cnt; ++i) {            dp[i] = str[i].g;            for(int j = 0; j < i; ++j)                if(str[i].c > str[j].c && str[i].k > str[j].k)                    dp[i] = max(dp[i], dp[j] + str[i].g);        }        int ans = -INF;        for(int i = 0; i < cnt; ++i)            ans = max(ans, dp[i]);        printf("Case %d: maximum height = %d\n", ++kase, ans);    }    return 0;}