uva 437 The Tower of Babylon（动态规划 嵌套矩形问题最长路）

有思路就去做，要相信自己

多处理更复杂的情况，你就不觉得现在复杂了

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;struct soli{    ll a,b,c;}s[40];int n;ll d[40][3];int vis[40][3];ll answer[40][3];ll dp(int s1,int s2){    if(vis[s1][s2])        return d[s1][s2];    vis[s1][s2]=1;    ll& ans=d[s1][s2];    ll temp1,temp2;    if(s2==0){temp1=s[s1].b;temp2=s[s1].c;ans=s[s1].a;}    if(s2==1){temp1=s[s1].a;temp2=s[s1].c;ans=s[s1].b;}    if(s2==2){temp1=s[s1].a;temp2=s[s1].b;ans=s[s1].c;}    ll temp=ans;    for(int i=1;i<=n;i++)    {        for(int j=0;j<=2;j++)        {            if(j==0)            {                if((s[i].b<temp1&&s[i].c<temp2)||(s[i].b<temp2&&s[i].c<temp1))                {                    ans=max(ans,dp(i,j)+temp);                }            }            if(j==1)            {                if((s[i].a<temp1&&s[i].c<temp2)||(s[i].a<temp2&&s[i].c<temp1))                {                    ans=max(ans,dp(i,j)+temp);                }            }            if(j==2)            {                if((s[i].b<temp1&&s[i].a<temp2)||(s[i].b<temp2&&s[i].a<temp1))                {                    ans=max(ans,dp(i,j)+temp);                }            }        }    }    return ans;}int main(){    int kase=0;    while(scanf("%d",&n)!=EOF)    {        if(n==0) break;        kase++;        ll a1,a2,a3;        memset(s,0,sizeof(s));        for(int i=1;i<=n;i++)        {            scanf("%I64d%I64d%I64d",&a1,&a2,&a3);            s[i].a=a1;            s[i].b=a2;            s[i].c=a3;        }        memset(answer,0,sizeof(answer));        for(int i=1;i<=n;i++)        {            for(int j=0;j<=2;j++)            {                memset(d,0,sizeof(d));                memset(vis,0,sizeof(vis));                answer[i][j]=dp(i,j);            }        }        ll max_=0;        for(int i=1;i<=n;i++)        {            for(int j=0;j<=2;j++)            {                max_=max(max_,answer[i][j]);            }        }        printf("Case %d: maximum height = %I64d\n",kase,max_);    }    return 0;}