HDU1069(DAG)

最简单的DAG了，只需要排个序，dp【i】 代表以第i个为结尾可以堆成的最大高度。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <stack>#include <queue>#include <list>#include <cmath>using namespace std;#define cir(a,b) memset(a,b,sizeof a)typedef long long  LL;const int maxn = 10000+10;const int INF = 1e9+10;int n;struct Brick{    int l,w,h;}s[maxn];int dp[maxn];bool cmp(const Brick& t1,const Brick& t2){    if(t1.l==t2.l) return t1.w<t2.w;    else return t1.l<t2.l;}int kase;int main(){    kase = 1;    while(cin >> n && n)    {        cir(dp,0);        int E = 1;        int l,w,h;        for(int i=0;i<10000;i++)        {            s[i].h = 0;        }        for(int i=1;i<=n;i++)        {            cin >> l >> w >> h;            s[E].l = l,s[E].w = w,s[E++].h = h;            s[E].l = l,s[E].w = h,s[E++].h = w;            s[E].l = w,s[E].w = l,s[E++].h = h;            s[E].l = w,s[E].w = h,s[E++].h = l;            s[E].l = h,s[E].w = l,s[E++].h = w;            s[E].l = h,s[E].w = w,s[E++].h = l;        }        //cout << "E : " << E <<endl;        sort(s+1,s+E+1,cmp);        int Max = -1;        for(int i=1;i<=E;i++) dp[i] = s[i].h;        for(int i=1;i<=E;i++)        {            for(int j=1;j<=i;j++)            {                if(s[i].l > s[j].l && s[i].w > s[j].w)                {                    dp[i] = max ( dp[i],dp[j]+s[i].h );                }            }            Max = max( dp[i],Max );            //cout << s[i].l << " " << s[i].w << " "  << s[i].h << " " <<endl;        }        printf("Case %d: maximum height = ",kase++);        cout << Max << endl;        //cout << H <<endl;    }    return 0;}